1
$\begingroup$

Let $L : \mathbb{R}^3\rightarrow\mathbb{R}^3$ be a linear transformation such that its matrix with respect to the standard basis is: $$[L] = \begin{bmatrix}1 & 0 & 1 \\ -1 & -1 & 0 \\0&1&-1\end{bmatrix}$$ Consider the two following bases of $\mathbb{R}^3$: $$\beta = \left(\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 1\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)$$ $$\gamma = \left(\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix},\begin{bmatrix}0 \\ 1 \\ 2\end{bmatrix},\begin{bmatrix}0 \\ 0 \\ 1\end{bmatrix}\right)$$ Find the matrix associated with the transformation with respect to $\beta$ in input and to $\gamma$ in output.

I seem to have some troubles with this kind of exercises. I have been explained a way to do it through multiplying $3$ matrices, but I can't seem to quite remember it every time I do an exercise. This is what I have tried:

The first thing is to rewrite the vectors in $\beta$ as linear combinations of those of the standard basis. Let's enumerate the vectors in $\beta$ and $\gamma$ with an index that goes from $1$ to $3$. Also, $e_1, e_2, e_3$ are the vectors of the standard basis of $\mathbb{R}^3$.

We observe that $\beta_1 = e_1 + e_2, \beta_2 = e_2+e_3, \beta_3 = e_3$, therefore the matrix that uses $\beta$ as the input basis and outputs vectors in the standard basis will be: $$[L]^{\beta}= \begin{bmatrix}1 & 1 & 1 \\ -2 & -1 & 0 \\0&0&-1\end{bmatrix}$$

Now, if my understanding is correct, I need to write the vectors that make up the columns of that matrix as linear combinations of vectors in $\gamma$.

Using Gauss' algorithm, we find that: $$\left[\begin{array}{ccc|ccc}1&0&0&1 & 1 & 1 \\ 2&1&0&-2 & -1 & 0 \\3&2&1&0&0&-1\end{array}\right]\rightarrow \left[\begin{array}{ccc|ccc}1 & 0 & 0 & 1 & 1 & 1 \\ 0 & 1 & 0 & -4 & -3 & -2 \\0 & 0 & 1 & 5 & 6 & 2\end{array}\right]$$

Which finally means that: $$[L]^{\beta}_{\gamma} = \left[\begin{array} & 1 & 1 & 1 \\ -4 & -3 & -2 \\5 & 6 & 2\end{array}\right]$$

Is my reasoning correct? Did I make any mistakes?

$\endgroup$

1 Answer 1

1
$\begingroup$

I would make it conceptually simpler:

Denote $P_\beta$ and $P_\gamma$ the change of basis matrices from the standard basis to the bases $\beta$ and $gamma$ respectively. These are the matrices with columns the coordinates of the vectors in $\beta$ and $gamma$ w.r.t. the standard basis.

Also, if $X$ is a column vector in the standard basis, denote $X_\beta$ and $X_\gamma$ the corresponding column vector in bases $\beta$ and $\gamma$. We know we have the relations. $$X=P_\beta X_\beta=P_\gamma X_\gamma.$$

Now, in the standard basis, the linear transformation is represented by the relation $Y=LX$, which becomes $$P_\gamma Y_\gamma=LP_\beta X_\beta,\quad\text{whence }\quad Y_\gamma=P_\gamma^{-1} LP_\beta X_\beta, $$ so that the matrix $L_{\beta\gamma}$ of the linear transformation, with the input in basis $\beta$ and the output in basis $\gamma$ is simply $$L_{\beta\gamma}=P_\gamma^{-1} LP_\beta.$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.