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Consider the linear transformation $T(M): \begin{bmatrix} 0&1\\0&0\end{bmatrix}M-M\begin{bmatrix} 0&1\\0&0\end{bmatrix}$ from $\mathbb{R}^{2\times2}$ to $\mathbb{R}^{2\times2}$.

a) What is the $\mathcal{B}$-matrix of the transformation $T$ with respect to the standard basis of $\mathbb{R}^{2\times2}$?

This part is easy for me, as we've been doing this over and over again at this part during the course. If $\big[[M]_{\mathcal{B}}\big]=\begin{pmatrix}a\\b\\c\\d\end{pmatrix}$, then $\big[[T(M)]_{\mathcal{B}}\big]=\begin{pmatrix}c\\d-a\\0\\c\end{pmatrix}$ such that $B=\begin{bmatrix}0&0&1&0\\-1&0&0&1\\0&0&0&0\\0&0&-1&0\end{bmatrix}$

b) Find bases of the image and kernel of $B$. This is the difficult part for me, despite having done this on its own a couple of times before. My first impulse was to put $B$ into row-reduced echelon form: $$rref(B)=\begin{bmatrix}1&0&0&-1\\0&0&1&0\\0&0&0&0\\0&0&0&0\end{bmatrix}$$

Okay, so we know that any solution of the system of linear equations $A\vec{x}=\vec{0}$ has $x_1=x_4$ and $x_2=0$. This will help us determine a basis of the kernel of $B$.

It's often easier to start with the image: $\vec{v_{1}}$ and $\vec{v_{3}}$ contain leading ones in $rref(B)$, so the corresponding (non-redundant) columns in $B$ form a basis of the image of $B$,$\therefore Im(A)=span \left\{\begin{pmatrix}0\\-1\\0\\0\end{pmatrix},\begin{pmatrix}1\\0\\0\\-1\end{pmatrix}\right\}$.

Okay, that's great. I know how to do this. But it's time-consuming. The textbook offers a more efficient method of computing this basis "by inspection" (how presumptuous!); I am comfortable doing this in some cases, but not in others. To save time on test day, I'd really like to get better at this method. The text states:

"Note that columns $\vec{v_{2}}$ and $\vec{v_{4}}$ are redundant, with $\vec{v_{2}}=\vec{0}$ and $\vec{v_{4}}=-\vec{v_{1}}$, or $\vec{v_{1}}+\vec{v_{4}}=\vec{0}$. Thus, the nonredundant columns $\vec{v_{1}}=\begin{bmatrix}0\\-1\\0\\0\end{bmatrix}, \vec{v_{3}}=\begin{bmatrix}1\\0\\0\\-1\end{bmatrix}$ form a basis of $Im(B),$ and $\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}1\\0\\0\\1\end{bmatrix}$ form a basis of $Ker(B).$"

Whoa. I clearly missed something about $\vec{0}$ in conceptualizing and defining a span of linearly independent vectors! What is it exactly about $\vec{0}$ that matters so much here? I figured that illustrating that a vector is a scalar multiple or additive function of any other vectors in a given span was enough to imply redundancy. What is so special about $\vec{0}$? Is it that the zero vector, by definition, is redundant? Given any nonzero vector $\vec{v}$, it would of course follow that $\vec{0}=\vec{v}+(-\vec{v})$, such that $\vec{0}$ is always a "redundant" vector in any given span. But this seems trivial.

As mentioned above, I know from the definition of the kernel (or, if you like, null space), that given a span of vectors in the target space of a linear transformation, the kernel is the subset of such a span such that all components of the subset satisfy $A\vec{x}=\vec{0}$. (please help me clarify the logic in my prose if it is at all ambiguous)

What is not crystal clear for me here is how we move from $x_1=x_4$ and $x_2=0$ to $Ker(B)=Span\left\{\begin{bmatrix}0\\1\\0\\0\end{bmatrix},\begin{bmatrix}1\\0\\0\\1\end{bmatrix}\right\}$. To be honest, I use a lot of the various online matrix calculators to do this when I'm doing my homework - but there are no calculators during the exam. The best logic I can come up with when constructing such a basis of a kernel is along these lines:

"Well, we know that the dimension of the image is 2, and the dimension of the $\mathcal{B}$-matrix is 4, so from rank-nullity we know that a basis of the kernel has to be two vectors...okay, so if $x_2=0$ as a solution to $A\vec{x}=\vec{0}$ then I have to put $\left\{(1,0)\right\}$ in each of the $x_2$-slots in the basis of the kernel..why? I DON'T KNOW! Because that's the answer?!...umm then we know that one of the vectors has to have $1's$ for the $x_1$- and $x_4$-slots because well, they equal each other... but we only have to illustrate this with one of the two vectors in the basis. And $x_3$ is always zero...and so that's how we get these two vectors..."

If anyone would be so kind as to help me delve deeper into the underlying logic behind constructing Bases of Kernels, here and in the general $n\times m$ case, I would greatly, greatly, appreciate it. If you're really up to the task, try explaining why I can't just put $Span\left\{\vec{0},\vec{0},\vec{0}\cdots \vec{0}\right\}$ as a basis of the kernel of any given linear transformation. Thank you!

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    $\begingroup$ You should try and shorten your question. The number of people on this site who are willing to tackle a wall of text like this is pretty small compared to the number of people are capable of answering questions about linear algebra. $\endgroup$ – Jim Mar 24 '15 at 3:26
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Let's start with:

$\begin{bmatrix}a\\b\\c\\d\end{bmatrix} \mapsto \begin{bmatrix}c\\d-a\\0\\-c\end{bmatrix}$

(You wrote this down slightly wrong, athough your $4 \times 4$ matrix is correct).

To find the nullspace of $T$, we need to know which values of $\{a,b,c,d\}$ yield the $0$-vector. It's clear that for this to happen, we must have $a = d$, and $c = 0$. Note that $b$ can be anything. This gives us two linearly independent vectors in the nullspace:

$\begin{bmatrix}1\\0\\0\\1\end{bmatrix}, \begin{bmatrix}0\\1\\0\\0\end{bmatrix}$.

In other words any vector in the nullspace is of the form:

$\begin{bmatrix}a\\b\\0\\a\end{bmatrix} = a\begin{bmatrix}1\\0\\0\\1\end{bmatrix} + b\begin{bmatrix}0\\1\\0\\0\end{bmatrix}$ which shows the two vectors above SPAN the nullspace.

In a similar vein, we can write any vector in $\text{Im }T$, as:

$c\begin{bmatrix}1\\0\\0\\-1\end{bmatrix} + (a-d)\begin{bmatrix}0\\-1\\0\\0\end{bmatrix}$, making it clear that these two linearly independent vectors span the image.

You seem to be VERY confused about the definition of linear independence, and linear dependence. A set $\{v_1,\dots,v_n\}$ of vectors is linearly independent if:

$c_1v_1 +\cdots + c_nv_n = 0 \implies c_1 = \cdots = c_n = 0$, that is, the ONLY linear combination of the $v_j$ that makes the $0$-vector is the $0$-combination.

A set is linearly dependent if there is SOME linear combination:

$c_1v_1 +\cdots + c_nv_n = 0$ with not all the $c_j = 0$. The set $\{0_V\}$ is always linearly dependent, since for ANY $c \neq 0$:

$c0_V = 0_V$

Bases must consist of linearly independent vectors, the $0$-vector is NEVER a basis vector. In this case, $x_2 = 0$ does NOT give a vector in the nullspace, in fact:

$T\left(\begin{bmatrix}1\\0\\1\\1\end{bmatrix}\right) = \begin{bmatrix}1\\0\\0\\-1\end{bmatrix} \neq 0$

I suspect your difficulty in finding the basis for $\text{ker }T$ in this example stems from the condition $c = 0$. The condition $c = 0$ means the nullspace lies within the subspace:

$\{(x_1,x_2,x_3,x_4) \in \Bbb R^4: x_3 = 0\}$ which has dimension $3$ (it has the basis:

$\{(1,0,0,0),(0,1,0,0),(0,0,0,1)\}$ -note all of these vectors have $0$ in the $3$rd coordinate. The absence of $b$ in the "typical" expression for $T(a,b,c,d)$ shows that:

$T(0,1,0,0) = (0,0-0,0,-0) = 0$. So:

$\langle (0,1,0,0)\rangle \subseteq \text{ker }T \subseteq \langle (1,0,0,0),(0,1,0,0),(0,0,0,1)\rangle$.

Note that if we restrict $T$ to acting on $\langle (1,0,0,0),(0,0,0,1)\rangle$ we have:

$T((a,0,0,0)+(0,0,0,d)) = (0,-a,0,0) + (0,d,0,0) = (0,d-a,0,0)$

and that is the $0$-vector if and only if $a = d$.

The vectors $\{(a,0,0,d) \in \Bbb R^4: a = d\}$ form a subspace of $\langle (1,0,0,0),(0,0,0,1)\rangle$, of dimension $1$, spanned by $\{(1,0,0,1)\}$.

Coupled with the above, we see that $T(\langle (1,0,0,1),(0,1,0,0)\rangle) = 0$.

Since $(1,0,0,1) \not\in \langle (0,1,0,0)\rangle$, and $T(1,0,0,0) = (0,-1,0,0) \neq 0$

we see that the set inclusions above are STRICT, so $1 < \dim(\text{ker }T) < 3$. The only number between $1$ and $3$ is $2$, and having found two LI vectors in the null-space, these must be a basis.

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