# If L : R n → R n is an invertible linear mapping, then [L] is also invertible (Is the standard matrix of an invertible matrix also invertible?

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

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

Thanks

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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.

mechanodroid
September 14, 2018 15:21 PM

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 $$[L^{-1}]=[L]^{-1}$$

egreg
September 14, 2018 15:42 PM

$L$ 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 $\mathbb R^n$).

Chris Custer
September 14, 2018 16:01 PM

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 $[L^{-1}][L] = I$.

copper.hat
September 14, 2018 17:41 PM