Let $F\left(\begin{matrix} a & b \\ c & d \end{matrix}\right) =\left(\begin{matrix} a & 0 \\ 0 & d \end{matrix}\right)$ be a linear function with $F : M(2\times2) \rightarrow M(2\times2)$.

  1. Which of the following are kernels and images of $F$? $$ a. \left(\begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix}\right) \\ b. \left(\begin{matrix} 0 & 4 \\ 2 & 0 \end{matrix}\right) \\ c. \left(\begin{matrix} 3 & 0 \\ 0 & -3 \end{matrix}\right) $$
  2. Describe kernel and image by specifying a base (for each)

My answer and thoughts

    • b is a kernel, because running it through $F$ yelds $0$.
    • 1. a and 1. c are images (but I'm not sure in case of 1.c, does this have to hold $b\neq c$? In case of 1.c they both are $0$)
    • I don't know how to write it out formally correct, but the kernel has to look like this $\left(\begin{matrix} 0 & x \\ y & 0 \end{matrix}\right)$ for any $x,y$.
    • I have no idea about how to write an image as a base.

For the kernel, I suspect something like this must hold

$$ \lambda_1\left(\begin{matrix} 0 & x \\ y & 0 \end{matrix}\right)+ \lambda_2\left(\begin{matrix} 0 & x \\ y & 0 \end{matrix}\right)= \left(\begin{matrix} 0 & 0 \\ 0 & 0 \end{matrix}\right) $$ which would mean that $x, y \neq 0$ for them to be linear independent.

  • $\begingroup$ For what matrix $m$, $F(m) = a$? This is what you should be asking when looking for an image. $\endgroup$ – Karolis Juodelė Nov 27 '12 at 14:20

I'll be assuming your matrices are real, but nothing much would change if they were complex.

The first thing to note is that the the kernel is a subset of the domain and the image is a subset of the codomain - this trips a lot of people up. To check if a matrix is in the kernel, you apply $F$ to it (like you correctly suggest for matrix $b$). However to check if a matrix $A$ is an "image", you need to see if there exists a matrix $M$ such that $F(M)=A$. So matrix $a$ is not an image, because every matrix in the image of $F$ has zero in the upper right and lower left. Matrix $c$ is an image, since

$$ F\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=\left(\begin{array}{cc}3 & 0\\ 0& -3\end{array}\right)=c $$

To find bases for the kernel and image, you can start with a base for $M(2\times 2)$. The standard base is the four matrices:

$$ e_1=\left(\begin{array}{cc}1 & 0\\ 0& 0\end{array}\right)\quad e_2=\left(\begin{array}{cc}0 & 1\\ 0& 0\end{array}\right)\quad e_3=\left(\begin{array}{cc}0 & 0\\ 1& 0\end{array}\right)\quad e_4=\left(\begin{array}{cc}0 & 0\\ 0& 1\end{array}\right) $$

Then, you know that any matrix $A\in M(2\times 2)$ can be written as $$ A=ae_1+be_2+ce_3+de_4. $$ Now apply $F$ to your basis elements: $$ F(e_1)=e_1\quad F(e_2)=0\quad F(e_3)=0\quad F(e_4)=e_4 $$

Since $F$ is linear, we can describe its range by taking an arbitrary element of $M(2\times 2)$, writing in terms of our base $\{e_1,e_2,e_3,e_4\}$, and applying $F$ to it: $$ F(ae_1+be_2+ce_3+de_4)=aF(e_1)+bF(e_2)+cF(e_3)+dF(e_4)=ae_1+de_4 $$

Thus the image can be described using the two basis matrices $e_1$ and $e_4$:

$$ \text{Im}F=\{\alpha e_1+\beta e_4:\alpha,\beta\in\Bbb{R}\} $$

Like you suggest, the kernel can be described as any matrix with zero diagonal. Now that we have a basis for $M(2\times 2)$, we can write this succinctly as

$$ \text{Ker}F=\{\alpha e_2+\beta e_3:\alpha,\beta\in\Bbb{R}\} $$

You should check to make sure these are subspaces.

Bonus question: show that any matrix $A\in M(2\times 2)$ can be written as the sum of a matrix in the kernel of $F$ and a matrix in the image of $F$.


For given $F : V → W$:

  1. $\operatorname{\ker}(F)=\{\,x\in V:F(x)=0\,\}$
  2. $\operatorname{im}(F)=\{\,w\in W:w=F(x),x\in V\,\}$

Kernel is a subspace of V, such that all of it elements $x$ give $F(x)=0$. In your case (b) is the right answer, because $F(b)=0$.

Image is a subspace of W, such that all of it elements $w$ are the result of some $F(x)=w$. So the answer is only (c).

The good answer on the second question has already been given.


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