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I am a tad confused about a couple of problems:

Find the standard matrix of the linear transformation $T(x, y, z) = (x − 2y + z, y − 2z, x + 3z)$.

Is this as simple as:

$$\begin{bmatrix} 1 & -2 & 1 \\ 0 & 1 & -2 \\ 1 & 0 & 3 \end{bmatrix}$$

The theorem I'm relying on is this:

enter image description here

enter image description here

In this proof, how does one get from the column of a's to the $T(ei)$ at the end?

  1. Let T be the reflection through the xz-coordinate plane in R3 : $T(x, y, z) = (x, −y, z)$. (a) Write the standard matrix A for the transformation T. (b) Use A to find the image of v = (1, −2, 2), and sketch both v and its image T(v).

a) $$\begin{bmatrix} 1 & -1 & 1 \end{bmatrix} = A$$ b) $A * (1,-2,2) = (1,2,2)$

Is this right?

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  • $\begingroup$ This seems right to me. $\endgroup$ – LordVader007 Sep 19 '18 at 4:13
  • $\begingroup$ The last-step equality of the proof relies on the assumption contained in the hypothesis. You assume that $T(e_i) = [a_{1i},\dotsc,a_{ni}]$ and this is used to prove that $T(v) = Av$. $\endgroup$ – Filippo De Bortoli Sep 19 '18 at 6:38
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  1. Yes, that is the right matrix. // One gets from the column of $a$'s to $T(e_i)$ at the end by applying the first equation in the theorem statement.

  2. (a) No, that is not right. In particular, since the matrix sends $\Bbb R^3\to \Bbb R^3$, it should be a $3\times 3$ matrix. (b) Yes, that is the right image, but of course it is not true that $(1,2,2)=Av$ using your matrix. In fact, using your $A$, $Av$ is not even defined, since your $A$ is a $3\times 1$ and $v$ is a $3\times 1$ also.

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