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The matrix $$ A=\left[\begin{array}{ccc} 1 & -a & a \\ -1 & a & a+2 \\ 1 & 2a+3 & -3a-4 \end{array}\right], $$ where $a \in \mathbb{R}$, represents a linear transformation $T: \mathbb{R}^3 \rightarrow \mathbb{R}^3$ with basis $$B=\{(1,0,-1),(1,-1,0),(1,1,1)\}.$$ It is known that $\dim \ker T=2$. Find $a$ and calculate $T(x,y,z)$ for all $(x,y,z)\in \mathbb{R}^3$

My solution:

Since $\dim\ker T=2$, I can conclude $\dim \mathrm{im}\, T=1$ since it is from $\mathbb{R}^3$ to $\mathbb{R}^3$.

I found that with $a=-1$, the matrix $A$ is equal to $$\left[\begin{array}{ccc} 1 & 1 & -1 \\ -1 & -1 & 1 \\ 1 & 1 & -1 \end{array}\right].$$ Therefore $\dim\ker T=2$ and $\dim \mathrm{im}\,T=1$.

Now I need to calculate $[T]_e$ using the columns: \begin{align} T(1,0,-1) & = 1(1,0,-1)-1(1,-1,0)+1(1,1,1) = (1,2,0), \\[0.1in] T(1,-1,0) & = 1(1,0,-1)-(1,-1,0)+1(1,1,1) = (1,2,0), \\[0.1in] T(1,1,1) & = -1(1,0,-1)+(1,-1,0)-1(1,1,1) = (-1,-2,0). \end{align} But $\dim\ker T=2$ and I get $\dim \mathrm{im}\,T=3$ instead of $1$.

Where did I go wrong?

Any ideas? Thanks.

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  • $\begingroup$ The only mistake you seem to have made up to this point is in saying that the dim(im T)=3; it does equal 1. $\endgroup$
    – user84413
    Commented Oct 8, 2014 at 17:48

1 Answer 1

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Notice that the matrix $A$ is relative to the basis $B$ and to find $T(x,y,z)$ you should find the matrix of $T$ relative to the canonical basis $B_c$. Let $P$ the change matrix from $B_c$ to $B$ i.e.

$$P=\begin{pmatrix}1&1&1\\0&-1&1\\-1&0&1\end{pmatrix}$$ then the matrix of $T$ relative to $B_c$ is

$$[T]_{B_c}=PAP^{-1}$$ hence $$T(x,y,z)=[T]_{B_c}(x,y,z)^T=PAP^{-1}(x,y,z)^T$$

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  • $\begingroup$ Can you please explain how did you reach to P? $\endgroup$
    – JaVaPG
    Commented Oct 8, 2014 at 23:09
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    $\begingroup$ The matrix $P$ is the three columns $(v_1\; v_2\; v_3)$ where the $v_i$'s are the three vectors of the basis $B$. $\endgroup$
    – user63181
    Commented Oct 8, 2014 at 23:12

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