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I was asked to calculate ImT and KerT for the following linear transformation. I got to a result for both, but as can be seen from my conclusions, I know there must be a mistake somewhere along the way. Yet I cannot find it… Can anyone help out?

The linear transformation $T:M_{2x2}(R) \to R_3[x]$ is defined by: $$T\begin{pmatrix}a&b\\c&d \end{pmatrix}$$$= (a-d)x^2+(b+c)x+5a-5d$

for every $$\begin{pmatrix}a&b\\c&d \end{pmatrix}$$$\in M_{2x2}(R)$

I reasoned in the following way:

For KerT:

a=d

b=-c

kerT: $$\begin{pmatrix}a&b\\-b&a \end{pmatrix}$$=$$a\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$+$$b\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$$

So KerT = Sp{$$\begin{pmatrix}1&0\\0&1 \end{pmatrix}$$,$$\begin{pmatrix}0&1\\-1&0 \end{pmatrix}$$} and as the matrices in KerT are clearly linearly independent, this is also a basis with dimension 2.

For ImT: $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp\{x^2,x,1\}$. This set is per definition linearly independent and therefore $Sp\{x^2,x,1\}$ is a basis to ImT and its dimension is 3.

The problem is that according to my logic, n = dimKerT + dimImT = 5 and this is contradictory to the fact that n = dimKerT + dimImT = 4, as the dimension of $M_{2x2}(R)=4$

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  • $\begingroup$ Use $\{ X\}$ for $\{ X\}$. $\endgroup$ – Shaun Jan 3 '19 at 12:38
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For ImT: $(a-d)x^2+(b+c)x+5a-5d$ is a linear combination of the standard basis of R3[x], namely $(x^2,x,1)$. So $ImT = Sp\{x^2,x,1\}$. This set is per definition linearly independent and therefore $Sp\{x^2,x,1\}$ is a basis to ImT and its dimension is 3.

It's not an arbitrary linear combination of those standard basis vectors, as the coefficients aren't independent. You could rewrite: $$(a-d)x^2+(b+c)x+5a-5d = (a-d)(x^2+5)+(b+c)x$$ so it can also be written as a linear combination of (only) $x^2+5$ and $x$, so...


Or for a more general approach, rewrite as (split in all the coefficients): $$(a-d)x^2+(b+c)x+5a-5d = a(\color{blue}{x^2+5})+b\color{red}{x}+c\color{purple}{x}+d(\color{green}{-x^2-5})$$ and reduce the spanning set $\left\{\color{blue}{x^2+5},\color{red}{x},\color{purple}{x},\color{green}{-x^2-5}\right\}$ to a basis by eliminating the linearly dependent elements.

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    $\begingroup$ Thank you! So my calculations for KerT were correct? $\endgroup$ – dalta Jan 3 '19 at 12:34
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    $\begingroup$ Yes, although you want to write a comma (instead of a $+$) between the two matrices that span the kernel (the basis has two elements, not the sum of two matrices); in the spanning set. $\endgroup$ – StackTD Jan 3 '19 at 12:35
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$$\text{Im}(T)\ne\text{Sp}\{1,x,x^2\}$$

This is because $T\Big(\begin{bmatrix}a&b\\c&d \end{bmatrix}\Big)=(a-d)(x^2+5)+(b+c)x=k_1(x^2+5)+k_2x$, which means $$\text{Im}(T)=\text{Sp}\{x,x^2+5\}$$ which is also a basis with dimension $2$.

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