Let the following be the linear transformation for $T$ from a $2$ x $2$ matrix to a second degree polynomial.

$T\begin{pmatrix}a&b\\ c&d\end{pmatrix}=\left(a-c\right)x^2+c\left(x-1\right)+b$

Find a basis for $Im(T)$ and $Ker(T)$.

So this is what I did to find the kernel.

I rearranged the equation to get:


So I tried to set everything in the brackets to $0$ and then got that:


This led me to think that a basis for $Ker(T)$ would be:

$\left\{\begin{pmatrix}0&0\\ 0&1\end{pmatrix}\right\}$

For the $Im(T)$, I rearranged the equation to get:


This gave me the following basis:


Which had the following matrix:

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

But all three columns of this matrix are linearly independent, which doesn't make any sense as the kernel I have calculated has dimension $1$.

Does anyone know where I have gone wrong here?

Thank you!


When you say, "Which has the following matrix", have a think about what you mean by this. You've taken a basis of polynomials (not a linear transformation) and somehow turned it into a matrix of numbers. What does this matrix mean? Why do you expect the columns to be linearly independent?

What you have done is form the change of basis matrix from the given basis to the standard basis, in the order $(x^2, x, 1)$. Such a matrix must certainly be invertible; change of basis matrices always are, and their inverse is the change of basis matrix from the second basis to the first.

Your issue seems to suggest that you're looking to find a matrix for the transformation, between the two bases. Bear in mind that this is impossible, since the basis is for the kernel and range, both of which are respectively subspaces of the domain and codomain, and the kernel happens to be a strict subspace. You would need full bases for the domain and codomain before you can compute a matrix for the transformation.

But, this isn't what the question wanted. It asked for two bases for the kernel and range, and you have found exactly that.

  • $\begingroup$ Oh yes I see. So would the basis of the image simply be the columns of the last matrix I found? $\endgroup$ – Dev SR Oct 12 '18 at 4:00
  • $\begingroup$ @DevSR yes exactly. I added some more information about how this is basically just a notational distinction between these spaces when you think of them as vector spaces. $\endgroup$ – CyclotomicField Oct 12 '18 at 4:00
  • $\begingroup$ Sorry I changed my comment right before you responded. So I just want to verify if the image is simply the 3 columns of the last matrix I found? Sorry for the inconvenience. I had mentioned the standard basis (row reduced form) before. $\endgroup$ – Dev SR Oct 12 '18 at 4:03
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    $\begingroup$ @DevSR The basis is the set of polynomials you found. The columns of the matrix are in $\mathbb{R}^3$, not $\mathbb{P}_2(\mathbb{R})$; they're in the wrong space. The columns are a representation of the basis in terms of a different basis: the standard basis (in a particular order). You can change the representation by changing the order, but ultimately, it's a coordinate representation of the basis, not the basis itself. $\endgroup$ – Theo Bendit Oct 12 '18 at 10:48
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    $\begingroup$ The basis for the kernel should be a subset of the domain, particularly the $2 \times 2$ matrices. You shouldn't represent it with numbers, so much as with matrices (or whatever the vectors are that make up your domain). In this case, the basis contains one vector (matrix), which you have found. Always, bases for the kernel are part of the domain, and bases for the range are part of the codomain! $\endgroup$ – Theo Bendit Oct 12 '18 at 15:22

You haven't gone wrong. You have a linear transformation from a four dimensional space to a three dimensional space. You can span the entire three dimensional space and still have a non-trivial kernel because you have a full dimension left over. Don't let the notation fool you, if you have a vector space of matrices column vectors or a polynomials you're essentially just changing where you put the indices. Matrix multiplication is what makes matrices special but as a linear space they behave like any other vector.

  • $\begingroup$ Oh yes I see. So will the basis for the image simply be the 3 columns of the last matrix I found? $\endgroup$ – Dev SR Oct 12 '18 at 4:08

Use a systematic approach. Let $$ E_1=\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix},\quad E_2=\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix},\quad E_3=\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix},\quad E_4=\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} $$ and let $p_1=1$, $p_2=x$, $p_3=x^2$. Then \begin{align} TE_1&=p_3\\ TE_2&=p_1\\ TE_3&=p_2-p_3\\ TE_4&=0 \end{align} so the matrix of $T$ with respect to the bases $\{E_1,E_2,E_3,E_4\}$ and $\{p_1,p_2,p_3\}$ is $$ A=\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 1 & 0 & -1 & 0 \end{pmatrix} $$ The RREF of $A$ is $$ U=\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$ which has rank $3$. Therefore the image of $T$ is the whole codomain.

The null space of $U$ (and of $A$) is generated by \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} which says that the kernel of $T$ is generated by $E_4$.

You can also go the hard way, but it's much simpler to observe that $$ p_1=TE_2,\quad p_2=T(E_3+E_2),\quad p_3=TE_1 $$ so the image is the whole space. Moreover $E_4$ clearly belongs to the kernel. By the rank-nullity theorem, the kernel has dimension $1$, so it is generated by $E_4$.


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