# mistake in calculations for ImT and KerT

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 $$M_{2x2}(R)=4$$

• Use $\{ X\}$ for $\{ X\}$. – Shaun Jan 3 '19 at 12:38

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.

• Thank you! So my calculations for KerT were correct? – dalta Jan 3 '19 at 12:34
• 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. – StackTD Jan 3 '19 at 12:35

$$\text{Im}(T)\ne\text{Sp}\{1,x,x^2\}$$

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 $$2$$.