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Let $v$ be an $n$-dimensional vector space over a field $F$ and $\psi: V \to V$ and isomorphism. Show that there exist bases $B_1$, $B_2$ (possibly different) such that the matrix representation of $\psi$ with respect to the two bases is precisely the $n \times n$ identity matrix.

I know that isomorphisms map bases to bases. I just cannot give the explicit two bases. I am assuming that one should fix a basis $\mathbf{v_1, v_2 \cdots v_n} $ and then use the fact that $\psi$ is an isomorphism to get the second basis. But I don't know how to proceed from here. Any suggestions?

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I think you already have it. If $f_j = \psi (v_j)$ then the matrix with respect to the $v_j$ and $f_j$ will be the identity. –  James S. Cook Mar 6 '13 at 4:24
    
Can you please show it more explicitly? Although it may be obvious, I am failing to see why. –  user44069 Mar 6 '13 at 4:26

3 Answers 3

Let $\{v_1,\ldots,v_n\}$ and $\{w_1,\ldots,w_n\}$ be any two bases for $V$. Recall that the matrix for a linear transformation $T:V\to V$ is given by $(a_{ij})$ where the entries $a_{ij}$ are determined by $$T(v_j)=a_{1j}w_1+\cdots a_{nj}w_n.$$ In other words, the matrix for $T$ looks like $$\left[T(v_1)\;\middle|\;\cdots \;\middle|\;\vphantom{$\strut_\strut^{\strut^\strut}$} T(v_n)\right]$$ Your goal is for this to be the identity matrix. Thus, you want $$T(v_1)=\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix}=1\cdot w_1+0\cdot w_2+\cdots +0\cdot w_n=w_1$$ $$\vdots$$ $$T(v_n)=\begin{bmatrix}0\\\vdots\\0\\1\end{bmatrix}=0\cdot w_1+0\cdot w_2+\cdots +1\cdot w_n=w_n$$ Thus, for any basis $B_1=\{v_1,\ldots,v_n\}$ of $V$ you want, letting the other basis $B_2$ just be $\{T(v_1),\ldots,T(v_n)\}$ will make the linear transformation $T$ be represented by the identity matrix.

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Let $v_1,v_2, \dots v_n$ be a basis for $V$. If $\psi: V \rightarrow V$ is an isomorphism then it maps a basis to a basis ( perhaps you need to prove this ?). So, define a new basis (possibly, assuming $\psi$ itself is not the identity) $f_j = \psi(v_j)$. Consider that $$ \psi (x_1v_1 + \cdots x_nv_n) = x_1f_1+ \cdots x_nf_n $$ Consequently the coordinate vector is unchanged as we map from domain to range and it follows the matrix which transports that vector is the identity. I can give more gory details if you wish...

In other notation, letting $\beta' =\{ f_1,\dots , f_n \}$ and $\beta = \{ v_1, \dots v_n \}$. If $y = x_1v_1 +\cdots + x_nv_n$ then $[y]_\beta = [x_1,\dots x_n]^T$ and $$ [\psi (y)]_{\beta'} = [x_1,\cdots x_n]^T $$ Thus $[\psi]_{\beta,\beta'} = I$.

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$\psi(v_i)$ will be the $i^{th}$ column of the matrix of $\psi$ with respect the basis $B=\{v_1, ...,v_n\}$ and the bases for the image $B^{\prime} = \{w_1, ..., w_n\}.$ For example:

If $\psi(x,y,z) = (2x + y, z, y+z)$ and you use the standard basis for the image and the domain then the columns of the matrix are:

$$\psi(e_1) = \psi(1,0,0) = (2, 0, 0) $$ $$\psi(e_2) = \psi(0,1,0) = (1,0,1) $$ $$\psi(e_3) = \psi(0,0,1) = (0,1,1)$$

So your matrix of $\psi$ is $$\left(\begin{array}{ccc} 2 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 1 \\ \end{array}\right)$$

If you pick your the basis for your image to be $B^{\prime} = \{w_1 = (2,0,0), w_2 = (1,0,1), w_3 = (0,1,1)\}$ then $$\psi(e_1) = \psi(1,0,0) = 1w_1 + 0w_2 + 0 w_3 = (1,0,0)$$ with respect to to the basis $B^{\prime}.$ Similarly, $$\psi(e_2) = \psi(0,1,0) = 0w_1 + 1w_2 + 0w_3 = (0,1,0)$$ with respect to $B^{\prime}.$

And finally you can verify that $\psi(e_3) = (0,0,1)$ with respect to $B^{\prime}$ thus the matrix of $\psi$ with respect to these two bases has the vector $(1,0,0)$ in the first column, $(0,1,0)$ in the second and $(0,0,1)$ in the third so it is the identity matrix, $$\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array}\right)$$

Try to take this example and see if you can make sense of the general case. Just map pick your new basis vectors so that that $w_i = \psi(v_i).$

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