0
$\begingroup$

We have $\phi\colon \mathbb{R}^4 \to \mathbb{R}^4, \phi(x_1,x_2,x_3,x_4)= (x_1-2x_3,-2x_3-x_4,x_1+2x_3+2x_4,x_1+x_4)$. We want to find the matrix of this transformation with respect to the following basis of $\mathbb{R}^4$:

\begin{align} \alpha_1&=(1,1,0,0)\\ \alpha_2&=(0,1,1,0)\\ \alpha_3&=(0,0,1,1)\\ \alpha_4&=(0,0,0,1)\end{align}

First, we can write down the transformation as $$ A_\phi = \begin{bmatrix} 1 & 0 & -2 & 0 \\ 0 & 0 & -2 & -1 \\ 1 & 0 & 2 & 2 \\ 1 & 0 & 0 & 1\\ \end{bmatrix} $$ and simplify it to the reduced row echelon form $$ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 0 & 1 & {1 \over 2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0\\ \end{bmatrix} $$ From this, we immediately see $\dim\ker\phi = 2$ and, by extension, $$\dim\operatorname{im} \phi = \dim\mathbb{R}^4 - \dim\ker \phi = 4 - 2 = 2.$$ So, we can choose whichever two linearly independent column vectors from $A_\phi$ and they will be a basis of $\operatorname{im}\phi$.

Now, how do we proceed from here to find the matrix of the transformation with respect to the given basis?

$\endgroup$
  • 1
    $\begingroup$ Calculate $\phi(\alpha_1)$. Express it as a linear combination of the $\alpha_i$: $\phi(\alpha_1)=a\alpha_1+b\alpha_2+c\alpha_3+d\alpha_4$. Then make $(a,b,c,d)$ the first column of a matrix. Repeat for the other $\alpha_i$ to get the other three columns of the matrix you have been asked to find. $\endgroup$ – Gerry Myerson Nov 29 '16 at 9:17
  • $\begingroup$ @GerryMyerson Thanks for the explanation. This is what I get from the procedure you have described. By taking $\phi(\alpha_i)$ we get the image of each $\alpha_i$. By finding the coefficients of the linear combination, we express these images in terms of the required basis. I am not quite clear why putting the coefficients in the columns of a matrix gets us what we want. $\endgroup$ – Zelazny Nov 29 '16 at 9:32
  • $\begingroup$ What you want is a matrix $A$ such that $\phi(v)=Av$ for all $v$ in ${\bf R}^4$, where $v$ and $Av$ are both described in terms of basis you've been given. With the matrix I described, you find, for example, $\phi(\alpha_1)=\phi(1,0,0,0)=A(1,0,0,0)=(a,b,c,d)$ which is what you want, since $\phi(\alpha_1)=a\alpha_1+b\alpha_2+c\alpha_3+d\alpha_4$. And similarly for the other $\alpha_i$, and then, by linearity, similarly for all $v$. $\endgroup$ – Gerry Myerson Nov 29 '16 at 11:45
  • 1
    $\begingroup$ I don’t understand why you row-reduced $A$. That and the conclusions you drew from it aren’t really relevant to this problem. $\endgroup$ – amd Nov 29 '16 at 19:54
  • 1
    $\begingroup$ Essentially what GerryMyerson described above in his comments. $\endgroup$ – amd Nov 30 '16 at 8:46
0
$\begingroup$

You don’t proceed from here. Finding the rank and nullity of a matrix doesn’t really tell you anything that you’d need to know to perform a change of basis on it.

Recall that the columns of a transformation matrix are the images of the domain basis vectors expressed relative to the basis of the codomain. So, for each basis vector $\alpha_j$, find the expression of $\phi\alpha_j$ as a linear combination $\sum_{i=1}^4 c_{ij}\alpha_i$. The required transformation matrix is then the coefficient matrix $[c_{ij}]$.

This computation can be accomplished all at once via matrix multiplication. Let $B$ be the matrix with the vectors $\alpha_k$ (expressed relative to the standard basis) as its columns. The columns of $A_\phi B$ are then the images of these vectors, also expressed relative to the standard basis. Observe that $B$ can be interpreted as converting from the $\{\alpha_k\}$ basis to the standard one (why?), so $B^{-1}$ converts from the standard basis to the $\{\alpha_k\}$ basis. Thus, $B^{-1}A_\phi B$ is the required matrix. This operation is known as a similarity transformation or conjugation of $A_\phi$.

$\endgroup$

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.