LAC Exercises (with solutions)

Section 2.7 Exercises (with solutions)

Exercises Exercises

1.

Let \(H\) be the subset of \(\R^4\) defined by

\begin{equation*} H = \left\{\ba{r}x_1\\x_2\\x_3\\x_4\ea : x_1 + x_2 + x_3 + x_4 = 0\right\}. \end{equation*}

Either show that \(H\) is a subspace of \(\R^4\text{,}\) or demonstrate how it fails to have a necessary property.

Solution.

The easiest way to show that \(H\) is a subspace is to note that it is the kernel of a linear map. Let \(A\) be the \(1\times 4\) matrix \(A =[1 \ 1\ 1\ 1]\text{.}\) Then

\begin{equation*} H = \{x \in \R^4\mid Ax=0\}, \end{equation*}

is the nullspace of \(A,\) which is always a subspace.

Alternatively of course you could check that 0 is in the set and that it is closed under addition and scalar multiplication.

2.

Suppose that \(T:\R^3\to \R^3\) is a linear map satisfying

\begin{equation*} T\left(\ba{r}3\\0\\0\ea\right) = \ba{r}6\\-3\\6\ea, T\left(\ba{r}1\\1\\0\ea\right) = \ba{r}2\\0\\1\ea, \text{ and } T\left(\ba{r}0\\0\\2\ea\right) = \ba{r}4\\6\\2\ea. \end{equation*}

(a)

If the standard basis for \(\R^3\) is \(\cE=\{e_1,e_2,e_3\},\) determine

\begin{equation*} T(e_1), T(e_2), \text{ and } T(e_3). \end{equation*}

Solution.

Using linearity, we are given \(T(3e_1) =3T(e_1)= \ba{r}6\\-3\\6\ea,\) so \(T(e_1)=\ba{r}2\\-1\\2\ea. \)

We are given \(T(e_1+e_2) = T(e_1)+ T(e_2)=\ba{r}2\\0\\1\ea,\) so

\begin{equation*} T(e_2) = T(e_1+e_2)-T(e_1) = \ba{r}2\\0\\1\ea - \ba{r}2\\-1\\2\ea = \ba{r}0\\1\\-1\ea. \end{equation*}

Finally, \(T(2e_3) = \ba{r}4\\6\\2\ea,\) so \(T(e_3) = \ba{r}2\\3\\1\ea.\)

(b)

Find \(T\left(\ba{r}1\\1\\1\ea\right).\)

Solution.

We compute

\begin{equation*} T\left(\ba{r}1\\1\\1\ea\right) = T(e_1)+T(e_2)+T(e_3) = \ba{r}4\\3\\2\ea. \end{equation*}

3.

Consider the upper triangular matrix

\begin{equation*} A = \ba{rrr}1\amp x\amp z\\0\amp 1\amp y\\0\amp0\amp1\ea, \end{equation*}

with \(x,y,z \in \R.\)

(a)

Give as many reasons as you can that shows the matrix \(A\) is invertible.

Solution.

We see that \(A\) is already in echelon (not RREF) form, which tells us there is a pivot in each column. Since there are only three variables the system \(Ax=0\) has only the trivial solution, to the linear map \(x\mapsto Ax\) is injective. Three pivots also means the column space is spanned by three independent vectors, so is all of \(\R^3.\) So the linear map is bijective, hence invertible.

One could also say that since the RREF of \(A\) is the identity matrix, it is invertible.

If you know about determinants, you could say the determinant equals 1, hence is nonzero, which means \(A\) is invertible.

(b)

Find the inverse of the matrix \(A.\)

Solution.

We row-reduce

\begin{align*} \ba{rrr|rrr}1\amp x\amp z\amp1\amp0\amp0\\ 0\amp 1\amp y\amp 0\amp 1\amp0\\ 0\amp 0\amp 1\amp 0\amp 0\amp 1\ea \amp\mapsto \ba{rrr|rrr}1\amp x\amp 0\amp 1\amp 0\amp -z\\ 0\amp 1\amp 0\amp 0 \amp 1 \amp -y\\ 0\amp0\amp1\amp0\amp0\amp1\ea\\ \amp\mapsto \ba{rrr|rrc} 1\amp0\amp0\amp1\amp -x\amp -z+xy\\ 0\amp 1\amp 0 \amp 0\amp 1\amp -y\\ 0\amp0\amp1\amp0\amp0\amp1\ea. \end{align*}

\begin{equation*} A^{-1} = \ba{rrc}1\amp -x\amp -z+xy\\ 0\amp 1\amp -y\\0\amp0\amp1\ea. \end{equation*}

4.

Consider the linear transformation \(T:\R^5\to \R^4\) given by \(T(x) = Ax\) where \(A\) and its reduced row-echelon form \(R\) are given by:

\begin{equation*} A=\ba{rrrrr} 1 \amp -1 \amp 2 \amp 6 \amp -3 \\ 2 \amp -1 \amp 0 \amp 7 \amp 10 \\ -2 \amp 3 \amp -7 \amp -15 \amp 17 \\ 2 \amp -2 \amp 2 \amp 8 \amp 5 \ea \text{ and } R= \ba{rrrrr} 1 \amp 0 \amp 0 \amp 5 \amp 0 \\ 0 \amp 1 \amp 0 \amp 3 \amp 0 \\ 0 \amp 0 \amp 1 \amp 2 \amp 0 \\ 0 \amp 0 \amp 0 \amp 0 \amp 1 \ea. \end{equation*}

(a)

Determine \(\ker T\text{,}\) the kernel of \(T.\)

Solution.

The kernel of \(T\) is the nullspace of \(A,\) which we know is the same as the nullspace of \(R\) which we can read off:

\begin{equation*} \ba{r} x_1\\x_2\\x_3\\x_4\\x_5 \ea= \ba{r} -5x_4\\-3x_4\\-2x_4\\x_4\\0 \ea =x_4\ba{r} -5\\-3\\-2\\1\\0 \ea \end{equation*}

(b)

Determine \(\Im T\text{,}\) the image of \(T.\)

Solution.

Depending upon what you already know, you could observe that the RREF \(R\) has a pivot in each row which means the columns of \(A\) span all of \(\R^4.\)

Or you may know that looking at \(R\) tells us there are four pivot columns in \(A\text{,}\) meaning the column space is spanned by 4 linearly independent vectors, hence the image is all of \(\R^4\text{.}\)

Or, if you have already learned the rank-nullity theorem, then from the previous part we would know the nullity is one, and so rank-nullity says the rank is \(5-1=4\text{,}\) so the image is a dimension 4 subspace of \(\R^4,\) which is all of \(\R^4.\)

5.

Let \(K\) be the set of solutions in \(\R^5\) to the homogeneous linear system

\begin{align*} x_1+x_2+x_3+x_4\phantom{+x_5}\amp =0\\ x_5\amp =0. \end{align*}

(a)

Find a basis \(\cB_0\) for \(K.\)

Solution.

The coefficient matrix for the system is

\begin{equation*} A = \ba{rrrrr}1\amp1\amp1\amp1\amp0\\0\amp0\amp0\amp0\amp1\ea \end{equation*}

which is already in reduced row-echelon form. We see there are two pivots, hence 3 free variables, meaning \(\dim K = 3.\) By inspection (or working out the details of finding all solutions), one finds a basis can be taken to

\begin{equation*} \cB_0 = \left\{v_1=\ba{r}-1\\1\\0\\0\\0\ea, v_2=\ba{r}-1\\0\\1\\0\\0\ea, v_3=\ba{r}-1\\0\\0\\1\\0\ea\right\}. \end{equation*}

(b)

Extend the basis \(\cB_0\) from the previous part to a basis \(\cB\) for all of \(\R^5.\)

Solution.

To extend a linearly independent set, one must add something not in the original span (see Theorem 3.1.3). There are many correct answers possible, but the vectors

\begin{equation*} v_4 = \ba{r}1\\1\\1\\1\\0\ea \text{ and } v_5=\ba{r}0\\0\\0\\0\\1\ea \end{equation*}

are clearly not in \(K\) since \(v_4\) does not satisfy the first defining equation, and \(v_5\) does not satisfy the second. So thinking algorithmically, \(\cB_0\cup \{v_4\}\) is linearly independent, and \(v_5\) is certainly not in the span of those four vectors since their last coordinates are all zero. Thus we may take (as one possible solution)

\begin{equation*} \cB=\cB_0\cup\{v_4,v_5\}. \end{equation*}

(c)

Define a linear transformation \(T:\R^5\to \R^5\) with kernel \(K\) and image equal to the set of all vectors with \(x_3=x_4=x_5=0.\)

Solution.

By Theorem 2.6.4, a linear map is uniquely defined by its action on a basis. It should be clear that the desired image is defined by the standard basis vectors \(e_1\) and \(e_2.\) So with the given basis \(\cB=\{v_1, \dots, v_5\},\) we must have

\begin{equation*} T(v_i) = 0, \text{ for }i=1,2,3\text{,} \end{equation*}

and \(T(v_4), T(v_5)\) linearly independent vectors in the image, say

\begin{equation*} T(v_4) = e_1\text{ and }T(v_5) = e_2. \end{equation*}

6.

Let \(M_{2\times 2}\) be the vector space of \(2\times 2\) matrices with real entries, and fix a matrix \(A=\ba{rr}a\amp b\\c\amp d\ea \in M_{2\times 2}\text{.}\) Consider the linear transformation \(T: M_{2\times 2} \to M_{2\times 2}\) defined by \(T(X) = AX\text{,}\) which (left) multiplies an arbitrary \(2\times 2\) matrix \(X\) by the fixed matrix \(A\text{.}\) Let \(\cE = \left\{ \e_1 = \ba{rr}1\amp 0\\0\amp 0\ea, \e_2 = \ba{rr}0\amp 1\\0\amp 0\ea, \e_3= \ba{rr}0\amp 0\\1\amp 0\ea, \e_4 = \ba{rr}0\amp 0\\0\amp 1\ea \right\}\) be a basis for \(M_{2\times 2}\text{.}\)

(a)

Find the matrix of \(T\) with respect to the basis \(\cE\text{,}\) that is \([T]_\cE\text{.}\)

Solution.

\begin{align*} T(\e_1)\amp = \ba{rr}a\amp b\\c\amp d\ea \ba{rr}1\amp 0\\0\amp 0\ea= \ba{rr}a\amp 0\\c\amp 0\ea= a\e_1 + c\e_3\\ T(\e_2)\amp = \ba{rr}a\amp b\\c\amp d\ea \ba{rr}0\amp 1\\0\amp 0\ea= \ba{rr}0\amp a\\0\amp c\ea= a\e_2 + c\e_4\\ T(\e_3)\amp = \ba{rr}a\amp b\\c\amp d\ea \ba{rr}0\amp 0\\1\amp 0\ea= \ba{rr}b\amp 0\\d\amp 0\ea= b\e_1 + d\e_3\\ T(\e_4)\amp = \ba{rr}a\amp b\\c\amp d\ea \ba{rr}0\amp 0\\0\amp 1\ea= \ba{rr}0\amp b\\0\amp d\ea= b\e_2 + d\e_4 \end{align*}

We now simply record the data as coordinate vectors:

\begin{equation*} [T]_\cE = \ba{rrrr}a\amp 0\amp b\amp 0\\0\amp a\amp 0\amp b\\c\amp 0\amp d\amp 0\\0\amp c\amp 0\amp d \ea \end{equation*}

(b)

Now let \(\cB\) be the basis, \(\cB=\{\e_1, \e_3, \e_2, \e_4\}\text{,}\) that is, the same elements as \(\cE\text{,}\) but with the second and third elements interchanged. Write down the appropriate change of basis matrix, \([I]_\cB^\cE\text{,}\) and use it to compute the matrix of \(T\) with respect to the basis \(\cB,\) that is \([T]_\cB.\)

Solution.

The change of basis matrices \([I]_\cB^\cE= \ba{rrrr}1\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 1\ea = [I]_\cE^\cB\text{,}\) so

\begin{align*} [T]_\cB\amp =[I]_\cE^\cB\,[T]_\cE\,[T]_\cB^\cE\\ \amp = \ba{rrrr}1\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 1\ea\ba{rrrr}a\amp 0\amp b\amp 0\\0\amp a\amp 0\amp b\\c\amp 0\amp d\amp 0\\0\amp c\amp 0\amp d\ea \ba{rrrr}1\amp 0\amp 0\amp 0\\0\amp 0\amp 1\amp 0\\0\amp 1\amp 0\amp 0\\0\amp 0\amp 0\amp 1\ea =\ba{rrrr}a\amp b\amp 0\amp 0\\c\amp d\amp 0\amp 0\\0\amp 0\amp a\amp b\\0\amp 0\amp c\amp d\ea. \end{align*}

Of course it was possible to write down \([T]_\cB\) simply from the information in part (a).

7.

Write down an explicit linear transformation \(T: \R^2 \to \R^3\) that has as its image the plane \(x-4y + 5z=0\text{.}\) What is the kernel of \(T\text{?}\)

Hint.

Any linear transformation \(T:\R^n \to \R^m\) has the form \(T(x) = Ax\) where \(A\) is the matrix for \(T\) with respect to the standard bases. How is the image of \(T\) related to the matrix \(A\text{?}\)

Solution.

We know that \(T\) can be given by \(T(x) = Ax\) where \(A\) is the \(3\times 2\) matrix whose columns are \(T(e_1)\) and \(T(e_2)\text{.}\) They must span the given plane, so for example, \(A = \ba{rr}4\amp -5\\1\amp 0\\0\amp 1\ea\) will do.

By rank-nullity, the kernel must be trivial.

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