Topics: the standard matrix of a linear transformation; geometric transformations of the plane; one-to-one and onto; transition matrices and difference equations.
By the end of this week you will be able to
Find the standard matrix of a linear transformation from a formula or from a geometric description.
Read off what a transformation does to the plane from its matrix, and the other way round.
Decide whether a linear transformation is one-to-one, onto, both, or neither, by counting pivot positions.
Build a transition matrix from a verbal description of how a population moves, and use it to step the population forward in time.
Explain what one-to-one and onto mean for such a model.
Day 1 & 2 — From a description of T to its matrix
A linear transformation is often handed to you in words or in pictures: "reflect every point through the line \(x_2 = x_1\)", "rotate the plane by 30 degrees". These days we turn any such description into a single matrix, and then read properties of the transformation off that matrix.
Facts we will use
A transformation \(T:\mathbb{R}^n \to \mathbb{R}^m\) is linear when \(T(\mathbf u + \mathbf v) = T(\mathbf u) + T(\mathbf v)\) and \(T(c\mathbf u) = c\,T(\mathbf u)\) for all \(\mathbf u, \mathbf v\) in \(\mathbb{R}^n\) and all scalars \(c\).
\(\mathbf e_1, \ldots, \mathbf e_n\) are the columns of the identity matrix \(I_n\). In \(\mathbb{R}^2\), \(\mathbf e_1 = \begin{bmatrix}1\\0\end{bmatrix}\) and \(\mathbf e_2 = \begin{bmatrix}0\\1\end{bmatrix}\).
The product \(A\mathbf x\) is the linear combination of the columns of \(A\) with weights taken from \(\mathbf x\).
Two examples first
Example 1
Let \(T:\mathbb{R}^2 \to \mathbb{R}^3\) be linear, and suppose all you are told is
Step 4. Recognise the answer as a matrix product. The last expression is a linear combination of two fixed vectors with weights \(x_1, x_2\), and that is exactly what \(A\mathbf x\) means when the two vectors are the columns of \(A\):
Check. Putting \(\mathbf x = \mathbf e_1 = (1,0)\) gives the first column, \((5,-7,2)\), which is the value we were given. Same for \(\mathbf e_2\).
Exercise 1
For self-checking only. This is not graded.
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be the linear transformation that reflects each point through the line \(x_2 = x_1\). Work out where \(\mathbf e_1\) and \(\mathbf e_2\) land, then enter the matrix \(A\) for which \(T(\mathbf x) = A\mathbf x\).
You may type decimals with a comma (1,5) or simple fractions (3/2) in any numeric box in this page.
Reflecting through the line \(x_2 = x_1\) swaps the two coordinates of a point, because the line is the set of points whose two coordinates agree. So \(\mathbf e_1 = (1,0)\) is sent to \((0,1) = \mathbf e_2\), and \(\mathbf e_2 = (0,1)\) is sent to \((1,0) = \mathbf e_1\).
The formula confirms the geometric description: the coordinates get swapped.
The method was general
Look back at what we actually used. In Example 1 the numbers 5, −7, 2, −3, 8, 0 played no role in the argument; neither did the fact that the target space was \(\mathbb{R}^3\). In Exercise 1 we never used a formula for the reflection, only the images of \(\mathbf e_1\) and \(\mathbf e_2\). The three steps were: split \(\mathbf x\) into \(x_1\mathbf e_1 + \cdots + x_n \mathbf e_n\), push \(T\) through, and collect the result into columns. That works for every linear transformation out of \(\mathbb{R}^n\).
Theorem 1 (standard matrix)
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be a linear transformation. There is exactly one matrix \(A\) with
\[ T(\mathbf x) = A\mathbf x \quad \text{for every } \mathbf x \text{ in } \mathbb{R}^n. \]
It is the \(m\times n\) matrix whose \(j\)th column is \(T(\mathbf e_j)\):
This matrix is called the standard matrix of \(T\). Two consequences are worth stating plainly. First, a linear transformation is completely determined by the \(n\) vectors \(T(\mathbf e_1), \ldots, T(\mathbf e_n)\): fix those, and every other value of \(T\) is forced. Second, "linear transformation" and "matrix transformation" describe the same collection of maps. The first name points at a property, the second at a way of computing.
Size check
If \(T:\mathbb{R}^n\to\mathbb{R}^m\), then \(A\) has \(n\) columns (one per input coordinate) and \(m\) rows (one per output coordinate). Getting this backwards is the most common slip in this section.
Transformations described geometrically
Example 2 (dilation)
Let \(T(\mathbf x) = 3\mathbf x\) on \(\mathbb{R}^2\). Then \(T(\mathbf e_1) = 3\mathbf e_1 = (3,0)\) and \(T(\mathbf e_2) = 3\mathbf e_2 = (0,3)\), so
\[ A = \begin{bmatrix}3 & 0\\ 0 & 3\end{bmatrix}. \]
Every point moves three times further from the origin along the same ray.
Example 3 (rotation)
Let \(T\) rotate every point of \(\mathbb{R}^2\) about the origin through an angle \(\varphi\), counterclockwise for positive \(\varphi\). We only need the two images. The point \(\mathbf e_1 = (1,0)\) sits on the unit circle at angle 0, so after rotating it sits at angle \(\varphi\), that is at \((\cos\varphi, \sin\varphi)\). The point \(\mathbf e_2\) sits at angle \(90^\circ\), so it lands at angle \(90^\circ + \varphi\), that is at \((-\sin\varphi, \cos\varphi)\). See Figure 1. Therefore
\[ A = \begin{bmatrix}\cos\varphi & -\sin\varphi\\ \sin\varphi & \cos\varphi\end{bmatrix}. \]
With \(\varphi = 90^\circ\) this is \(\begin{bmatrix}0 & -1\\ 1 & 0\end{bmatrix}\), which does send \((1,0)\) to \((0,1)\).
Figure 1. A rotation by \(\varphi = 40^\circ\). Both basic vectors keep their length 1 and turn by the same angle, so their images are \((\cos\varphi,\sin\varphi)\) and \((-\sin\varphi,\cos\varphi)\). These two images are the columns of the standard matrix.
Because a linear transformation of the plane is fixed by its two columns, it is fixed by what it does to the unit square with corners \((0,0), (1,0), (1,1), (0,1)\). Figure 2 shows six standard transformations this way. Reading such a picture is a fast way to guess a matrix, and reading a matrix is a fast way to predict the picture.
(a) \(\begin{bmatrix}1&0\\0&-1\end{bmatrix}\)
(b) \(\begin{bmatrix}-1&0\\0&1\end{bmatrix}\)
(c) \(\begin{bmatrix}1&1\\0&1\end{bmatrix}\)
(d) \(\begin{bmatrix}1&0\\0&2\end{bmatrix}\)
(e) \(\begin{bmatrix}1/2&0\\0&1\end{bmatrix}\)
(f) \(\begin{bmatrix}1&0\\0&0\end{bmatrix}\)
Figure 2. The unit square, dashed, and its image, shaded, under six linear transformations. In each panel the image of \(\mathbf e_1\) is the lower edge of the shaded region and the image of \(\mathbf e_2\) is the edge leaving the origin upward; those two vectors are the columns of the matrix listed under the panel. In panel (f) the square collapses onto a segment.
Example 4 (one transformation after another)
Let \(T\) first apply the horizontal shear that sends \(\mathbf e_2\) to \(\mathbf e_2 - 0.5\,\mathbf e_1\) and leaves \(\mathbf e_1\) alone, and then reflect the result through the \(x_2\)-axis. Find the standard matrix of \(T\).
Solution
We only have to follow the two basic vectors through both stages.
\(\mathbf e_1\): the shear leaves it at \(\mathbf e_1\). The reflection through the \(x_2\)-axis flips the sign of the first coordinate, so \(\mathbf e_1 \mapsto -\mathbf e_1 = (-1, 0)\).
\(\mathbf e_2\): the shear sends it to \(\mathbf e_2 - 0.5\mathbf e_1 = (-0.5,\,1)\). Reflecting flips the first coordinate again, giving \((0.5,\,1)\).
These two images are the columns:
\[ A = \begin{bmatrix}-1 & 0.5\\ 0 & 1\end{bmatrix}. \]
Nothing new was needed here. Composing transformations always works like this at our level: track \(\mathbf e_1\) and \(\mathbf e_2\) through every stage in order, then write the final positions as columns.
One-to-one and onto
The two questions we asked about linear systems in Chapter 1 have names once we speak about transformations. "Does a solution exist?" becomes "is \(T\) onto?", and "is the solution unique?" becomes "is \(T\) one-to-one?".
Definition (onto)
\(T:\mathbb{R}^n\to\mathbb{R}^m\) is onto\(\mathbb{R}^m\) if every \(\mathbf b\) in \(\mathbb{R}^m\) is the image of at least one\(\mathbf x\) in \(\mathbb{R}^n\). Equivalently, the equation \(T(\mathbf x) = \mathbf b\) has a solution for every \(\mathbf b\), so the range of \(T\) fills the whole codomain.
Definition (one-to-one)
\(T:\mathbb{R}^n\to\mathbb{R}^m\) is one-to-one if every \(\mathbf b\) in \(\mathbb{R}^m\) is the image of at most one\(\mathbf x\) in \(\mathbb{R}^n\). Equivalently, \(T(\mathbf x) = \mathbf b\) has either exactly one solution or none.
The two conditions are independent: a transformation can have either, both, or neither. Figure 3 shows the difference.
Figure 3. Onto is about the range covering the codomain. One-to-one is about no two different inputs sharing an image. The bottom row shows a failure of one-to-one: two distinct vectors with the same image.
Example 5 (a transformation that is neither)
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) be the projection onto the \(x_1\)-axis, panel (f) of Figure 2, with matrix \(A = \begin{bmatrix}1&0\\0&0\end{bmatrix}\), so \(T(x_1,x_2) = (x_1, 0)\).
Solution
Not one-to-one: the two different vectors \((2,3)\) and \((2,-5)\) both have image \((2,0)\). One \(\mathbf b\) with two preimages is enough to break the condition.
Not onto \(\mathbb{R}^2\): every image has second coordinate 0, so \(\mathbf b = (0,1)\) is never reached. The range is the \(x_1\)-axis, a line inside the plane.
Checking every \(\mathbf b\) by hand is not a method. The next two statements turn both questions into facts about the columns of \(A\), which we already know how to test. We state them without proof; the examples after them show how they are used.
Theorem 2
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be linear. Then \(T\) is one-to-one if and only if the equation \(T(\mathbf x) = \mathbf 0\) has only the trivial solution \(\mathbf x = \mathbf 0\).
Theorem 3
Let \(T:\mathbb{R}^n\to\mathbb{R}^m\) be linear with standard matrix \(A\). Then:
\(T\) maps \(\mathbb{R}^n\) onto \(\mathbb{R}^m\) if and only if the columns of \(A\) span \(\mathbb{R}^m\);
\(T\) is one-to-one if and only if the columns of \(A\) are linearly independent.
Both parts reduce to counting pivot positions in an echelon form of \(A\), where \(A\) is \(m\times n\):
Question
Condition on \(A\)
Pivot test
Is \(T\) onto \(\mathbb{R}^m\)?
columns span \(\mathbb{R}^m\)
a pivot in every row (\(m\) pivots)
Is \(T\) one-to-one?
columns independent
a pivot in every column (\(n\) pivots)
Since a matrix cannot have more pivots than it has rows, or more than it has columns, the shape alone already rules things out. A transformation \(\mathbb{R}^n\to\mathbb{R}^m\) can only be onto when \(n \ge m\), and can only be one-to-one when \(n \le m\).
Does \(T\) map \(\mathbb{R}^4\) onto \(\mathbb{R}^3\)? Is \(T\) one-to-one?
Solution
\(A\) is already in echelon form, so the pivots can be read directly: they sit in positions \((1,1)\), \((2,2)\) and \((3,4)\), entries 1, 2 and 5.
Onto: there are 3 pivots and 3 rows, so every row has one. By the table, \(T\) is onto \(\mathbb{R}^3\): whatever \(\mathbf b\) in \(\mathbb{R}^3\) you name, \(A\mathbf x = \mathbf b\) is consistent.
One-to-one: there are 4 columns but only 3 pivots, so column 3 has no pivot and \(x_3\) is a free variable. Then \(A\mathbf x = \mathbf 0\) has a nontrivial solution, and by Theorem 2 \(T\) is not one-to-one.
This matches the shape rule: from \(\mathbb{R}^4\) to \(\mathbb{R}^3\) we have \(n = 4 > 3 = m\), so one-to-one was impossible from the start.
Example 7
Let \(T(x_1,x_2) = (3x_1 + x_2,\; 5x_1 + 7x_2,\; x_1 + 3x_2)\). Show that \(T\) is linear and one-to-one. Is it onto \(\mathbb{R}^3\)?
Solution
Write the output as a matrix product by reading off the coefficient of \(x_1\) and of \(x_2\) in each coordinate:
A transformation given by a matrix product is linear, so \(T\) is linear with the standard matrix \(A\) shown.
One-to-one: the two columns \((3,5,1)\) and \((1,7,3)\) are not multiples of each other, so they are linearly independent. By Theorem 3(b), \(T\) is one-to-one.
Onto:\(A\) is \(3\times 2\), so it has at most 2 pivots while it has 3 rows. Some row has no pivot, the columns cannot span \(\mathbb{R}^3\), and \(T\) is not onto \(\mathbb{R}^3\). Its range is the plane through the origin spanned by the two columns.
Exercise 2
For self-checking only. This is not graded.
Build a \(3\times 2\) matrix \(A\) so that the transformation \(\mathbf x \mapsto A\mathbf x\) from \(\mathbb{R}^2\) to \(\mathbb{R}^3\) is one-to-one but not onto \(\mathbb{R}^3\). Many different matrices work, and any correct one will be accepted.
Any \(3\times 2\) matrix whose two columns are not multiples of each other works, for instance
Its columns are independent, so the transformation is one-to-one by Theorem 3(b). It cannot be onto \(\mathbb{R}^3\), because 2 columns give at most 2 pivots for 3 rows. In fact the "not onto" part is automatic: every\(3\times2\) matrix fails to be onto \(\mathbb{R}^3\), so the only real task was making the columns independent.
Day 3 — Transition matrices and difference equations
Today we use the same matrix in a different way. Instead of applying it once to a geometric object, we apply it once per time step to a vector that describes the state of a population, and watch how the state moves.
Facts we will use
For a \(2\times 2\) matrix, \(\begin{bmatrix}a & b\\ c& d\end{bmatrix}\begin{bmatrix}p\\q\end{bmatrix} = \begin{bmatrix}ap + bq\\ cp + dq\end{bmatrix}\): each output entry is one row of the matrix paired with the vector.
For a \(2\times 2\) matrix \(A\), the first column is \(A\mathbf e_1\) and the second column is \(A\mathbf e_2\).
If two columns of a square matrix are equal, they are linearly dependent.
The set-up
Suppose a fixed population is divided into a few groups, and we record how many members are in each group once per period: once a year, once a month, once a day. Collect the counts in a vector. With two groups,
\[ \mathbf x_k = \begin{bmatrix}\text{size of group 1 at time } k\\ \text{size of group 2 at time } k\end{bmatrix}. \]
This is called the state vector at time \(k\). Now suppose that during each period a fixed fraction of each group moves to each of the other groups, and that these fractions do not change from period to period. For example, with two groups it might be that every year 10% of group 1 moves to group 2 while the other 90% stays put, and 4% of group 2 moves to group 1 while the other 96% stays put. The word "fixed" is doing real work here: the same four percentages are used in year 1, in year 2 and in every later year, even though the group sizes they act on keep changing. Then the state at time \(k+1\) can be computed from the state at time \(k\) by one matrix product:
An equation of this form is a difference equation, and \(M\) is the transition matrix (also called a migration matrix). Given the starting state \(\mathbf x_0\), you get \(\mathbf x_1\), then \(\mathbf x_2\), and so on, each from the previous one.
How to read the entries
The entry in row \(i\), column \(j\) of \(M\) is the fraction of group \(j\) that ends up in group \(i\) after one period. So columns are "from" and rows are "to". Column \(j\) describes what happens to the members of group \(j\), and since all of them go somewhere, each column adds up to 1. That last fact is worth checking every time you write a transition matrix down; it catches most mistakes immediately.
Example 8 (city and suburbs)
In a certain region, each year about 5% of the city's residents move to the suburbs and the other 95% stay, while about 3% of the suburban residents move into the city and the other 97% stay. In 2020 the city had 600,000 residents and the suburbs 400,000. Find the populations in 2021 and 2022.
Solution
Step 1. Choose the order of the groups. Take group 1 to be the city and group 2 the suburbs, so
Step 2. Build the matrix one column at a time. Column 1 says where this year's city residents are next year: 95% of them in the city, 5% in the suburbs, so column 1 is \((0.95,\, 0.05)\). Column 2 says where this year's suburban residents are next year: 3% in the city, 97% in the suburbs, so column 2 is \((0.03,\, 0.97)\). Hence
\[ M = \begin{bmatrix}0.95 & 0.03\\ 0.05 & 0.97\end{bmatrix}. \]
Both columns add to 1, as they must. See Figure 4.
The first entry is \(570{,}000 + 12{,}000 = 582{,}000\): the city keeps 570,000 of its own people and gains 12,000 from the suburbs. The second is \(30{,}000 + 388{,}000 = 418{,}000\). So in 2021 the split is
Step 5. Check the total. Each state vector adds up to 1,000,000. That is not a coincidence: the columns of \(M\) each add to 1, so nobody is created or lost, only moved. If your totals drift, you have a wrong entry.
Figure 5 continues the same computation for ten years.
Figure 4. The movement described in Example 8. The two arrows leaving each circle carry the fractions of that group, so they add to 1. Those two numbers are one column of \(M\): the pair 0.95 and 0.05 leaving the city is column 1.Figure 5. Applying \(M\) from Example 8 ten times in a row. The city falls from 600,000 to about 472,700 and the suburbs rise to about 527,300, with the crossing happening in year 7. Both curves flatten toward the dashed levels 375,000 and 625,000; why they settle there is the subject of a later chapter.
Example 9 (three groups: subscription tiers)
A service has three plans: Free, Plus and Pro. Each month, of the Free users, 80% stay on Free, 15% move to Plus and 5% move to Pro. Of the Plus users, 10% drop to Free, 70% stay and 20% move up to Pro. Of the Pro users, 5% drop to Free, 10% drop to Plus and 85% stay. This month there are 10,000 Free, 4,000 Plus and 1,000 Pro users. Find next month's numbers.
Solution
Step 1. Fix the order as Free, Plus, Pro, so \(\mathbf x_0 = (10{,}000,\; 4{,}000,\; 1{,}000)\).
Step 2. One column per starting plan. The Free sentence gives column 1, the Plus sentence column 2, the Pro sentence column 3, each listed in the order Free, Plus, Pro:
So \(\mathbf x_1 = (8{,}450,\; 4{,}400,\; 2{,}150)\), and the total is still 15,000.
Step 4. Repeat if you want another month. Applying \(M\) to \(\mathbf x_1\) gives \(\mathbf x_2 = (7{,}307.5,\; 4{,}562.5,\; 3{,}130)\). The model produces fractions of a person; round only when you report the numbers, not in the middle of the computation.
Link back to Day 1 and 2. The map \(\mathbf x \mapsto M\mathbf x\) is a linear transformation from \(\mathbb{R}^3\) to \(\mathbb{R}^3\), and \(M\) is its standard matrix. Theorem 1 says the columns are \(M\mathbf e_1, M\mathbf e_2, M\mathbf e_3\), and here that reads: column 1 is next month's split of a population that starts out entirely on Free. That is exactly how we wrote the columns down in Step 2.
Can the model be run backwards?
Suppose you are handed this month's state \(\mathbf x_1\) and asked what last month's state \(\mathbf x_0\) was. You are asking whether \(M\mathbf x = \mathbf x_1\) has exactly one solution, which is precisely the question "is \(\mathbf x\mapsto M\mathbf x\) one-to-one?" from Day 1 and 2. And asking which states could ever occur one step from now is the question "is it onto?".
Both have columns adding to 1, so both are legitimate transition matrices for two groups. Is the past recoverable in each case?
Solution
The first model. The columns \((0.9, 0.1)\) and \((0.2, 0.8)\) are not multiples of one another, so they are linearly independent. By Theorem 3(b) the transformation is one-to-one, so at most one \(\mathbf x_0\) produces a given \(\mathbf x_1\): last month's split is uniquely determined. Two independent columns in a \(2\times2\) matrix also span \(\mathbb{R}^2\), so by Theorem 3(a) the transformation is onto as well.
The second model. The two columns of \(M_2\) are equal, hence linearly dependent, so the transformation is not one-to-one. It is easy to see this directly. Take a population of 100 people:
Two completely different starting states give the same result, so seeing \((60,40)\) tells you nothing about where those people started. The reason is visible in the matrix: both columns say "60% to group 1, 40% to group 2" regardless of the starting group, so after one step the original split has been erased.
It is not onto either. Every image is \(x_1(0.6,0.4) + x_2(0.6,0.4) = (x_1+x_2)(0.6,0.4)\), a multiple of a single vector, so the range is the line spanned by \((0.6,0.4)\) rather than all of \(\mathbb{R}^2\). A state such as \((100, 0)\) can never be reached.
Exercise 3
For self-checking only. This is not graded.
A campus bike share has two stations, North and South. Each day 30% of the bikes that started at North end the day at South, and the other 70% end at North. Of the bikes that started at South, 20% end at North and 80% end at South. This morning there are 120 bikes at North and 80 at South.
Take group 1 to be North and group 2 to be South. Enter the transition matrix \(M\) and tomorrow morning's state \(\mathbf x_1\).
\(M = \)
\(\mathbf x_1 = \)
Column 1 describes the bikes starting at North: 0.70 stay at North, 0.30 go to South. Column 2 describes the bikes starting at South: 0.20 go to North, 0.80 stay at South. So
Write down a \(2\times2\) transition matrix for two groups, so that all entries are between 0 and 1 and each column adds to 1, and so that yesterday's state cannot be recovered from today's. Many different matrices work.
"Yesterday's state cannot be recovered" means the transformation is not one-to-one, which by Theorem 3(b) means the two columns are linearly dependent. For a \(2\times2\) matrix that means one column is a multiple of the other. Combined with the requirement that both columns add to 1, the multiple has to be 1, so the two columns must be equal. Any matrix of the form
works, for example \(\begin{bmatrix}0.5 & 0.5\\ 0.5 & 0.5\end{bmatrix}\) or \(\begin{bmatrix}1 & 1\\ 0 & 0\end{bmatrix}\). In words: the destination of a member does not depend at all on the group it came from, so after one step the original split is gone.
Homework — Week 4
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Four questions, 100 points in total.
Question 1 — Standard matrices20 points
(a) (10 points)
Let \(T:\mathbb{R}^2\to\mathbb{R}^3\) be linear with \(T(\mathbf e_1) = (3, 0, -2)\) and \(T(\mathbf e_2) = (1, 4, 5)\). Write the standard matrix of \(T\), then compute \(T(2,-1)\), showing the combination you used.
(b) (10 points)
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) first reflect each point through the \(x_1\)-axis and then multiply every coordinate by 2. Find the standard matrix of \(T\) by following \(\mathbf e_1\) and \(\mathbf e_2\) through both stages, and say in one sentence what the second column of your matrix means.
Question 2 — Two grocery chains25 points
A small town has two grocery chains, Ada and Bay. Each month 20% of Ada's customers switch to Bay and the other 80% stay with Ada, while 30% of Bay's customers switch to Ada and the other 70% stay with Bay. This month Ada has 4,000 customers and Bay has 6,000. Take group 1 to be Ada and group 2 to be Bay.
(a) (5 points)
Write the transition matrix \(M\), and state in one sentence what its second column means.
(b) (10 points)
Compute the state vectors \(\mathbf x_1\) and \(\mathbf x_2\) for the next two months. Write out the products, not only the answers.
Final answers:
\(\mathbf x_1 = \)
\(\mathbf x_2 = \)
(c) (4 points)
The total number of customers is the same in every month. Explain why, using a property of \(M\).
(d) (6 points)
Is the transformation \(\mathbf x\mapsto M\mathbf x\) one-to-one? Justify your answer with one of this week's theorems, then say what your answer means for someone who knows this month's split and wants to work out last month's.
Row reduce the augmented matrix \([\,A \mid \mathbf 0\,]\) to echelon form. Start by filling in the first grid, then add one row operation and one grid per step.
(c) (8 points)
Using your echelon form, decide whether \(T\) is one-to-one and whether it is onto \(\mathbb{R}^3\). Name the pivot positions you are counting and the theorem you are using.
(d) (4 points)
Describe the set of all \(\mathbf x\) with \(T(\mathbf x) = \mathbf 0\) in parametric vector form, and say what kind of geometric object it is in \(\mathbb{R}^3\).
Question 4 — Composition and shape25 points
(a) (15 points)
Let \(T:\mathbb{R}^2\to\mathbb{R}^2\) first rotate each point counterclockwise about the origin by \(90^\circ\) and then project the result onto the \(x_1\)-axis. Find the standard matrix of \(T\) by following \(\mathbf e_1\) and \(\mathbf e_2\) through both stages. Then decide whether \(T\) is one-to-one and whether it is onto \(\mathbb{R}^2\), and describe the range of \(T\) as a geometric object.
(b) (10 points)
Let \(S:\mathbb{R}^5\to\mathbb{R}^3\) be linear, with standard matrix \(B\). For each of the following, say whether it is possible and justify your answer by counting pivot positions in \(B\).