by Tim Weah
Last Updated September 14, 2018 15:20 PM

I'm having difficulty with this practice question. Please give me any hints or give me a place to start.

Thanks

If $L$ is invertible and $\{e_1, \ldots, e_n\}$ is the canonical basis for $\mathbb{R}^n$, then $\{Le_1, \ldots, Le_n\}$ is also a basis for $\mathbb{R}^n$.

By definition, the columns of $[L]$ are precisely $\{Le_1, \ldots, Le_n\}$. Hence $\operatorname{rank} [L] = n$ so $[L]$ is invertible.

More generally, if $V$ is a vector space with basis $\mathscr{B}=\{v_1,\dots,v_n\}$, we can define the *coordinate* map $C_{\mathscr{B}}\colon V\to\mathbb{R}^n$ where
$$
C_{\mathscr{B}}(v)=\begin{bmatrix}\alpha_1\\\alpha_2\\ \vdots \\ \alpha_n\end{bmatrix}
\quad\text{if and only if}\quad
v=\alpha_1v_1+\alpha_2v_2+\dots+\alpha_nv_n
$$
It is readily seen that $C_{\mathscr{B}}$ is well defined, linear and bijective.

Now suppose $W$ is another vector space with basis $\mathscr{D}$.

Theorem.If $L\colon V\to W$ is a linear map, then there exists a unique matrix $A$ such that, for all $v\in V$,$$ C_{\mathscr{D}}(L(v))=AC_{\mathscr{B}}(v) $$

Such matrix $A$ is sometimes denoted as $[L]_{\mathscr{B}}^{\mathscr{D}}$ or variations thereof.

The uniqueness stated in the theorem allows us to easily prove that if $Z$ is another vector space with basis $\mathscr{F}$, and $M\colon W\to Z$ is a linear map, then $$ [M\circ L]_{\mathscr{B}}^{\mathscr{F}}= [M]_{\mathscr{D}}^{\mathscr{F}} [L]_{\mathscr{B}}^{\mathscr{D}}\tag{*} $$

Among the basic properties of $[L]_{\mathscr{B}}^{\mathscr{D}}$ is that its rank is the same as the rank of $L$ (dimension of the image). In particular, if $L$ is invertible, then also $[L]_{\mathscr{B}}^{\mathscr{D}}$ is invertible.

Your situation is even easier, because you have $V=\mathbb{R}^n$ and $\mathscr{B}=\mathscr{D}$ is the standard (canonical) basis. Then, omitting to specify the bases in the associated matrix, the relation (*) becomes (where $\iota$ denotes the identity map $$ [\iota]=[L^{-1}\circ L]=[L^{-1}][L] $$ and it is obvious that $[\iota]$ is the identity matrix. Therefore $$ [L^{-1}]=[L]^{-1} $$

$L$ is invertible $\iff [L]_{\mathcal B}$ for any basis $\mathcal B$ is invertible $\iff [L]_{\mathcal S}$ where $\mathcal S$ is the standard basis is invertible.

The easiest way to see it is that we are talking in any case about the *same* linear transformation (so it does the same thing to vectors in $\mathbb R^n$).

I'm not exactly sure what you mean by standard matrix, I presume you mean the matrix with respect to some fixed basis $b_1,...,b_n$ in the domain and codomain. Let $c$ denote the coordinates with respect to this basis, that is, $c(\sum_k x_k b_k ) = (x_1,...,x_n)^T$. Note that $c$ is linear and invertible, with $c^{-1}((x_1,...,x_n)^T) = \sum_k x_k b_k $.

Then we have $[L]x = c(L(c^{-1}(x))$ (this is how $[L]$ is defined).

To show that $[L]$ is invertible, it is sufficient to show that it is injective.

Suppose $[L]x = 0$, then $c(L(c^{-1}(x)) = 0$ and hence $L(c^{-1}(x)) = 0$. Since $L$ is invertible, we have $c^{-1}(x) = 0$ and so $x=0$.

Alternatively, note that $[L^{-1}][L]x = c(L^{-1}(c^{-1}(c(L(c^{-1}(x)))))) = c(L^{-1}(L(c^{-1}(x))) = c(c^{-1}(x)) =x $, and so $[L^{-1}][L] = I$.

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