Show that $T$ is a linear transformation and find a basis for $N(T)$

Question

Consider the vector space of 2-by-2 matrices $M_{2\times 2}(\mathbb{R})$ over $\mathbb{R}$. For a given matrix A=$\begin{pmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{pmatrix}$, consider the map $T:M_{2\times 2}(\mathbb{R}) \to \mathbb{R}$ given by $T(A)=a_{2,1}$

$a)$ Show that T is a linear transformation.

Definition: Let $V$ and $W$ be vector spaces over $F$. We call a function $T : V \to W$ a linear transformation from $V$ to $W$ if, for all $x, y \in V$ and $c \in F$, we have

$(1) T (x + y) = T (x) + T (y)$

$(2) T (cx) = cT (x)$

Using this definition this is the solution I came up with

$T(c\begin{pmatrix}a_{1,1} &a_{1,2}\\a_{2,1}&a_{2,2}\end{pmatrix}+\begin{pmatrix}b_{1,1} &b_{1,2}\\b_{2,1}&b_{2,2}\end{pmatrix}) =c(a_{2,1})+(b_{2,1})=cT(A)+T(B)$

$b)$ Find a basis for $N(T)$

Definition: Null Space $N(T )=\{x \in V : T (x) = 0\}$.

Not sure where to begin with this. The following questions I assume can only be solved if finding a solution to $b)$

$c)$ Calculate the dimension of $N(T)$.

$d)$ Find a basis for $R(T)$

Thanks in advance.

Answer 1

You solution to part (a) seems to be just fine. For part (b):

Consider what is needed for a matrix $A\in\mathbb{R}^{(2,2)}$ to map to $0$ under $T$, i.e. $T(A)=0$, i.e. $a_{2,1}=0$. Thus

$$ N(T)=\{A\in\mathbb{R}^{(2,2)}\mid a_{2,1}=0\}=\{\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}\in\mathbb{R}^{(2,2)}\} $$

The classical basis for $\mathbb{R}^{(2,2)}$ is the basis associated with the canonical basis for $\mathbb{R}^4$:

$$ \begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\1 &0\end{pmatrix}, \begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$

You may check that these are indeed linear independent and span $\mathbb{R}^{(2,2)}$. Coming back to the null space of $T$, you can see that every matrix $A\in N(T)$, i.e. every matrix of the form

$$ \begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix} $$

can be created as a linear combination of

$$ \begin{pmatrix}1 &0\\0 &0\end{pmatrix}, \begin{pmatrix}0 &1\\0 &0\end{pmatrix},\begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$

with

$$ \begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix}=a_{1,1}\begin{pmatrix}1 &0\\0 &0\end{pmatrix} + a_{1,2} \begin{pmatrix}0 &1\\0 &0\end{pmatrix}+ a_{2,2}\begin{pmatrix}0 &0\\0 &1\end{pmatrix} $$

as the defining condition for $N(T)$ is a zero at position $a_{2,1}$. As we found a set of linear independent vectors(the three matrices above) that span $N(T)$, we've found a basis for the concerned space. Followingly, as an answer to part (c), we find that $\mathrm{dim}(N(T))=3$.

Note, that by the rank-nullity theorem(and assuming that $R(T)$ means the range of $T$), we have that

$$ 4=\mathrm{dim}(\mathbb{R}^{(2,2)})=\mathrm{dim}(N(T))+\mathrm{dim}(R(T))=3+\mathrm{dim}(R(T)) $$

Therefore, $\mathrm{dim}(R(T))=1$, i.e. as $T:\mathbb{R}^{(2,2)}\to\mathbb{R}$, we have $R(T)=\mathbb{R}$. A typical basis for $\mathbb{R}$ as a vector space is the number $1$.

Answer 2

Let $A_1= \begin{pmatrix}1 &0\\0 &0\end{pmatrix}$, $A_2= \begin{pmatrix}0 &1\\0 &0\end{pmatrix}$ and $A_3= \begin{pmatrix}0 &0\\0 &1\end{pmatrix}$, then

$A \in N(T) \iff A=\begin{pmatrix}a_{1,1} &a_{1,2}\\0 &a_{2,2}\end{pmatrix} \iff $

$A \in span \{A_1,A_2,A_3\} $.

Can you proceed ?

Answer 3

a. Let $X=[x_{i,j}],Y=[y_{i,j}]$ be 2x2 matrices and $c\in F$. (1) $T(X+Y) = x_{2,1} + y_{2,1} = T(X)+T(Y)$ (2)$T(cX) = cx_{2,1} = cT(X). $ This proves that T is a linear trasformation.

b. $N(T) = \left\lbrace A = [a_{i,j}] | a_{2,1}=0 \right\rbrace = Span\left( \left\lbrace \left[ \begin{array}{cc} 1&0\\0&0\end{array}\right] , \left[ \begin{array}{cc} 0&1\\0&0\end{array}\right],\left[ \begin{array}{cc} 0&0\\0&1\end{array}\right] \right\rbrace \right)$

Show that $\left\lbrace \left[ \begin{array}{cc} 1&0\\0&0\end{array}\right] , \left[ \begin{array}{cc} 0&1\\0&0\end{array}\right],\left[ \begin{array}{cc} 0&0\\0&1\end{array}\right] \right\rbrace$ is linearly independent to how that it is a basis.