Exercise set D

Please, see the general comment on the tutorial exercises

Question D.1

Consider the matrix $\mathbf{A}$ defined by

$$\begin{array}{r}\mathbf{A}=\left(\begin{array}{cc}1& 0\\ 0.5& -12\\ -2& 7\end{array}\right)\end{array}$$

Do the columns of this matrix form a basis of ${\mathbb{R}}^3$? Why or why not?

Question D.2

Is ${\mathbb{R}}^2$ a linear subspace of ${\mathbb{R}}^3$? Why or why not?

Question D.3

Show that if $T:{\mathbb{R}}^K\to {\mathbb{R}}^N$ is a linear function then $\mathbf{0}\in \ker (T)$.

Question D.4

Let $S$ be any nonempty subset of ${\mathbb{R}}^N$ with the following two properties:

• $\mathbf{x},\mathbf{y}\in S⟹\mathbf{x}+\mathbf{y}\in S$

• $c\in \mathbb{R}$ and $\mathbf{x}\in S⟹c\mathbf{x}\in S$

Is $S$ a linear subspace of ${\mathbb{R}}^N$?

Question D.5

If $S$ is a linear subspace of ${\mathbb{R}}^N$ then any linear combination of $K$ elements of $S$ is also in $S$. Show this for the case $K=3$.

Question D.6

Let $\{{\mathbf{x}}_1,{\mathbf{x}}_2\}$ be a linearly independent set in ${\mathbb{R}}^2$ and let $\gamma$ be a nonzero scalar. Is it true that $\{\gamma {\mathbf{x}}_1,\gamma {\mathbf{x}}_2\}$ is also linearly independent?

Question D.7

Is

$$\begin{array}{r}z=\left(\begin{array}{c}-3.98\\ 11.73\\ -4.32\end{array}\right)\end{array}$$

in the span of $X:=\{{\mathbf{x}}_1,{\mathbf{x}}_2,{\mathbf{x}}_3\}$, where

$$\begin{array}{r}{\mathbf{x}}_1=\left(\begin{array}{c}-4\\ 0\\ 0\end{array}\right),{\mathbf{x}}_2=\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right),{\mathbf{x}}_3=\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)?\end{array}$$

Question D.8

What is the rank of the $N\times N$ identity matrix $\mathbf{I}$?

What about the upper-triangular matrix which non-zero elements are $1$?

Question D.9

Show that if $T:{\mathbb{R}}^N\to {\mathbb{R}}^N$ is nonsingular, i.e. linear bijection, the inverse map $T^{-1}$ is also linear.

Solutions

Question D.1

No, these two vectors do not form a basis of ${\mathbb{R}}^3$. If it did then ${\mathbb{R}}^3$ would be spanned by just two vectors. This is impossible. For example, it would imply by the exchange lemma that any three vectors in ${\mathbb{R}}^3$ are linearly dependent. We know this is false.

Question D.2

This is a bit of a trick question, but to solve it you just need to look carefully at the definitions (as always). A linear subspace of ${\mathbb{R}}^3$ is a subset of ${\mathbb{R}}^3$ with certain properties. ${\mathbb{R}}^3$ is a collection of 3-tuples $(x_1,x_2,x_3)$ where each $x_i$ is a real number. Elements of ${\mathbb{R}}^2$ are 2-tuples (pairs), and hence not elements of ${\mathbb{R}}^3$. Therefore ${\mathbb{R}}^2$ is not a subset of ${\mathbb{R}}^3$, and in particular not a linear subspace of ${\mathbb{R}}^3$.

Question D.3

Let $T$ be as in the question. We need to show that $T\mathbf{0}=\mathbf{0}$. Here’s one proof. We know from the definition of scalar multiplication that $0\mathbf{x}=\mathbf{0}$ for any vector $\mathbf{x}$. Hence, letting $\mathbf{x}$ and $\mathbf{y}$ be any vectors in ${\mathbb{R}}^K$ and applying the definition of linearity,

$$T\mathbf{0}=T(0\mathbf{x}+0\mathbf{y})=0T\mathbf{x}+0T\mathbf{y}=\mathbf{0}+\mathbf{0}=\mathbf{0}$$

Question D.4

Yes, $S$ must be a linear subspace of ${\mathbb{R}}^N$. To see this, pick any $\mathbf{x}$ and $\mathbf{y}$ in $S$ and any scalars $\alpha ,\beta$. To establish our claim we need to show that $\mathbf{z}:=\alpha \mathbf{x}+\beta \mathbf{y}$ is in $S$. To see that this is so observe that by ($\text{ such that }ar\text{ such that }ar$) we have $\mathbf{u}:=\alpha \mathbf{x}\in S$ and $\mathbf{v}:=\beta \mathbf{y}\in S$. By ($\text{ such that }ar$) we then have $\mathbf{u}+\mathbf{v}\in S$. In other words, $\mathbf{z}\in S$ as claimed.

Question D.5

Let ${\mathbf{x}}_i\in S$ and ${\alpha }_i\in \mathbb{R}$ for $i=1,2,3$. We claim that

(9)$${\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2+{\alpha }_3{\mathbf{x}}_3\in S$$

To see this let $\mathbf{y}:={\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2$. By the definition of linear subspaces we know that $\mathbf{y}\in S$. Using the definition of linear subspaces again we have $\mathbf{y}+{\alpha }_3{\mathbf{x}}_3\in S$. Hence (9) is confirmed.

Question D.6

The answer is yes. Here’s one proof: Suppose to the contrary that $\{\gamma {\mathbf{x}}_1,\gamma {\mathbf{x}}_2\}$ is linearly dependent. Then one element can be written as a linear combination of the others. In our setting with only two vectors, this translates to $\gamma {\mathbf{x}}_1=\alpha \gamma {\mathbf{x}}_2$ for some $\alpha$. Since $\gamma \ne 0$ we can multiply each side by $1/\gamma$ to get ${\mathbf{x}}_1=\alpha {\mathbf{x}}_2$. But now each ${\mathbf{x}}_i$ is a multiple of the other. This contradicts linear independence of $\{{\mathbf{x}}_1,{\mathbf{x}}_2\}$.

Here’s another proof: Take any ${\alpha }_1,{\alpha }_2\in \mathbb{R}$ with

(10)$${\alpha }_1\gamma {\mathbf{x}}_1+{\alpha }_2\gamma {\mathbf{x}}_2=\mathbf{0}$$

We need to show that ${\alpha }_1={\alpha }_2=0$. To see this, observe that

$${\alpha }_1\gamma {\mathbf{x}}_1+{\alpha }_2\gamma {\mathbf{x}}_2=\gamma ({\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2)$$

Hence $\gamma ({\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2)=\mathbf{0}$. Since $\gamma \ne 0$, the only way this could occur is that ${\alpha }_1{\mathbf{x}}_1+{\alpha }_2{\mathbf{x}}_2=\mathbf{0}$. But $\{{\mathbf{x}}_1,{\mathbf{x}}_2\}$ is linearly independent, so this implies that ${\alpha }_1={\alpha }_2=0$. The proof is done.

Question D.7

There is an easy way to do this: We know that any linearly independent set of 3 vectors in ${\mathbb{R}}^3$ will span ${\mathbb{R}}^3$. Since $\mathbf{z}\in {\mathbb{R}}^3$, this will include $\mathbf{z}$. So all we need to do is show that $X$ is linearly independent. To this end, take any scalars ${\alpha }_1,{\alpha }_2,{\alpha }_3$ with

$$\begin{array}{r}{\alpha }_1\left(\begin{array}{c}-4\\ 0\\ 0\end{array}\right)+{\alpha }_2\left(\begin{array}{c}1\\ 2\\ 0\end{array}\right)+{\alpha }_3\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right)=\mathbf{0}:=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\end{array}$$

Write as a linear system of 3 equations and show that the only solution is ${\alpha }_1={\alpha }_2={\alpha }_3=0$. In this case the set would be linearly independent.

Question D.8

By definition, $\mathrm{rank}(\mathbf{I})$ is equal to the dimension of the span of its columns. Its columns are the $N$ canonical basis vectors in ${\mathbb{R}}^N$, which we know span all of ${\mathbb{R}}^N$. Hence

$$\mathrm{rank}(\mathbf{I})=\dim ({\mathbb{R}}^N)=N$$

Draft of the proof for the second question: For the upper triangular matrix start by showing that the columns are linearly independent, and because there are $N$ of them, they span the whole space ${\mathbb{R}}^N$, thus the expression above applies again, and the rank is $N$.

Question D.9

Let $T:{\mathbb{R}}^N\to {\mathbb{R}}^N$ be nonsingular and let $T^{-1}$ be its inverse. To see that $T^{-1}$ is linear we need to show that for any pair $\mathbf{x},\mathbf{y}$ in ${\mathbb{R}}^N$ (which is the domain of $T^{-1}$) and any scalars $\alpha$ and $\beta$, the following equality holds:

$$T^{-1}(\alpha \mathbf{x}+\beta \mathbf{y})=\alpha T^{-1}\mathbf{x}+\beta T^{-1}\mathbf{y}.$$

In the proof we will exploit the fact that $T$ is by assumption a linear bijection.

So pick any vectors $\mathbf{x},\mathbf{y}\in {\mathbb{R}}^N$ and any two scalars $\alpha ,\beta$. Since $T$ is a bijection, we know that $\mathbf{x}$ and $\mathbf{y}$ have unique preimages under $T$. In particular, there exist unique vectors $\mathbf{u}$ and $\mathbf{v}$ such that

$$T\mathbf{u}=\mathbf{x}\text{and}T\mathbf{v}=\mathbf{y}$$

Using these definitions, linearity of $T$ and the fact that $T^{-1}$ is the inverse of $T$, we have

$$\begin{array}{r}{T}^{-1}(\alpha \mathbf{x}+\beta \mathbf{y})=T^{-1}(\alpha T\mathbf{u}+\beta T\mathbf{v})\\ =T^{-1}(T(\alpha \mathbf{u}+\beta \mathbf{v}))\\ =\alpha \mathbf{u}+\beta \mathbf{v}\\ =\alpha T^{-1}\mathbf{x}+\beta T^{-1}\mathbf{y}.\end{array}$$

This chain of equalities confirms

$$T^{-1}(\alpha \mathbf{x}+\beta \mathbf{y})=\alpha T^{-1}\mathbf{x}+\beta T^{-1}\mathbf{y}.$$