# Find the matrix of the transformation with respect to the given basis

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?

• 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. – Gerry Myerson Nov 29 '16 at 9:17
• @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. – Zelazny Nov 29 '16 at 9:32
• 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 $v$. – Gerry Myerson Nov 29 '16 at 11:45
• I don’t understand why you row-reduced $A$. That and the conclusions you drew from it aren’t really relevant to this problem. – amd Nov 29 '16 at 19:54
• Essentially what GerryMyerson described above in his comments. – amd Nov 30 '16 at 8:46

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 $A_\phi$.