Given the Matrix for the Linear Transformation T. Calculate T(v)

Question

I have been given the matrix

$$A = \begin{pmatrix} 1 & 8 & -17 \\ 3 & 2 & -2 \\ -3 & -6 & 9 \\ 1 & -4 & 7 \\ \end{pmatrix} $$

which is the matrix of the linear transformation T. I have also been given the vector

$$v = \begin{pmatrix} 3 \\ 0 \\ -3 \\ \end{pmatrix} $$

and am supposed to calculate T(v). Overall, I'm a bit confused on how to solve this honestly, but I'm thinking about somehow making row reductions for [A v], but in order to do this I'll have to transpose A to get the dimensions right, does that make any sense or have I gotten it all wrong?

Any input is very much appreciated!

Thanks, Nick

Answer 1

The product $A\cdot v$ is a new vector the coordinates thereof are the dot products of the rows of $A$ by the vector $v$.

For instance, the first coordinate of $A\cdot v$ is $$1\cdot 3+8\cdot 0 -17\cdot (-3)= 54.$$

Similarly, the coefficients of the product of an $m{\times} n$ matrix $A$ by an $n{\times}p$ matrix $B$ are the dot products of the rows of $A$ by the columns of $B$.

Answer 2

You don't need to transpose A in order to compute it, the dimensions are fine.

By definition, the product AB is defined for $A_{n\times k},B_{k\times l}$, and in your case, the dimensions of A are $4\times 3$ and the vector is $3\times 1$, so the product is well defined.

A is the matrix representing T, so $T(v)$ is $Av$ (in the right basis), So $T(v)$ is: $$ \begin{pmatrix} 1 & 8 & -17\\ 3 & 2 & -1\\ -4 & -6 & 9\\ 1 & -4 & 7 \end{pmatrix}\begin{pmatrix} 3\\ 0\\ -3 \end{pmatrix} = \begin{pmatrix} 1\cdot3+8\cdot0-17\cdot(-3)\\ 3\cdot3+2\cdot0-1\cdot(-3)\\ -4\cdot3-6\cdot0+9\cdot(-3)\\ 1\cdot3-4\cdot0+7\cdot(-3) \end{pmatrix} = \begin{pmatrix} 54\\ 12\\ -39\\ -18 \end{pmatrix} $$

So your answer (in the basis the matrix is given at) is $$ \begin{pmatrix} 54\\ 12\\ -39\\ -18 \end{pmatrix} $$