# Finding invertible matrices that solve a given equation

I need to work out the following question(as homework) but have no clue how to go about solving the problem. Really, no idea. I'd appreciate if someone could point out what steps are necessary in order to solve the problem.

$f:\mathbb{R^5} \to \mathbb{R^4}$ with the following matrix relative to the unit bases $E_5$ and $E_4$.

$M = \begin{bmatrix}-2&0&0&0&-2\\0&1&-2&0&0\\-2&0&0&1&1\\0&1&-2&1&3\end{bmatrix}$

Determine the invertible matrices $S$ and $T$ such that:

$M= S \times M(f) \times T^{-1} = \begin{bmatrix}1&0&\cdots&0\\0&\ddots& &\\\vdots&&1\\ & &&0\\0&\cdots&0&0\end{bmatrix}$

• Hint: find bases for the image and kernel of $M$. – amd Jun 25 '17 at 19:39
• Funnily enough, that's the next question. – Charmaine N Ndolo Jun 25 '17 at 19:41
• Also, use the fact that the columns of the matrix are the images of the basis. – amd Jun 25 '17 at 19:42
• Actually, it’s not just any basis for the image of $M$, but the images of a basis for a complement of $\ker M$ that you need. – amd Jun 25 '17 at 19:57

This problem essentially wants you to find bases for $\mathbb R^5$ and $\mathbb R^4$ such that the matrix of $f$ has the desired form. Recalling that the columns of a transformation matrix are the images of the domain basis vectors, we can see that the last few columns get mapped to zero, so the corresponding basis vectors of $\mathbb R^5$ must span $\ker f$. Similarly, the leading columns tell us that the images of the other basis vectors are a basis for $\operatorname{im}f$.

So, start by computing a basis for the kernel (null space) of $f$ and extend that to a complete basis of $\mathbb R^5$. $T$ will be the inverse of the matrix with these vectors as its columns. For $S$, compute $Mv$ for each vector $v$ of the basis extension that you just computed. This will be a basis for the image of $f$. Extend that to cover all of $\mathbb R^4$. Again, $S$ will be the inverse of the matrix with these vectors as its columns.

Note that the matrices $S$ and $T$ aren’t unique. You can freely choose any basis for the kernel, any extension of it, and any extension of the image basis.

You can proceed row-reducing $M$: $$\begin{bmatrix}-2&0&0&0&-2\\0&1&-2&0&0\\-2&0&0&1&1\\0&1&-2&1&3\end{bmatrix} \to \begin{bmatrix}1&0&0&0&1\\0&1&-2&0&0\\0&0&0&1&3\\0&0&0&0&0\end{bmatrix}.$$ The rref gives you almost everything you need for the solution. You can read a basis for the null space of $M$ from the rref: $(0,2,1,0,0)$ and $(1,0,0,3,-1)$. The non-zero rows of the rref also give you a basis for the row space, which is the orthogonal complement of the null space. Right-multiplying $M$ by the matrix with these vectors as its columns gives $$\begin{bmatrix}-2&0&0&0&-2\\0&1&-2&0&0\\-2&0&0&1&1\\0&1&-2&1&3\end{bmatrix}\begin{bmatrix}1&0&0&1&0\\0&1&0&0&-2\\0&-2&0&0&-1\\0&0&1&3&0\\1&0&3&-1&0\end{bmatrix} = \begin{bmatrix}-4&0&-6&0&0\\0&5&0&0&0\\-1&0&4&0&0\\3&5&10&0&0\end{bmatrix}.$$ The columns of this matrix are the images of the domain basis vectors. The non-zero columns are a basis for the image of $M$, so take them and extend to a complete basis of the codomain with any linearly-independent vector: $(0,0,0,1)^T$ will do here. Collecting these vectors into a matrix and left-multiplying by its inverse will produce the required matrix. The problems want $SMT^{-1}$, though, so $S$ and $T$ will be the inverses of the two matrices that you’ve constructed.

You could also proceed as in Omnomnomnom’s now-deleted answer. Observe that the first, second and fourth columns of the rref of $M$ already have the required form. Right-multiplying a matrix by the $j$th column of the identity (i.e., the $j$th standard basis vector) picks out its $j$th column, so start building the right-hand matrix by taking $\mathbf e_1$, $\mathbf e_2$ and $\mathbf e_4$ as its first three columns. The rref also tells you that the third column of $M$ is $-2$ times the second and the last column is a linear combination of the first and fourth. Since the columns of a matrix product are linear combination of the columns of the left-hand factor, you can use this information to generate zeros in the last two columns of the product: $$\begin{bmatrix}-2&0&0&0&-2\\0&1&-2&0&0\\-2&0&0&1&1\\0&1&-2&1&3\end{bmatrix}\begin{bmatrix}1&0&0&0&1\\0&1&0&-2&0\\0&0&0&-1&0\\0&0&1&0&3\\0&0&0&0&-1\end{bmatrix} = \begin{bmatrix}-2&0&0&0&0\\0&1&0&0&0\\-2&0&1&0&0\\0&1&1&0&0\end{bmatrix}.$$ From here you can proceed as above to construct the left-hand matrix, and again, you’ll need to invert both to get $S$ and $T$. (In this case, inverting the right-hand matrix is quite simple: it will be a permutation of the matrix that you’ve built.)

• Thanks a lot for your help. I followed your steps, I'm not sure if I have it right though. One thing I don't understand is how you can read the basis of the null space of M from the rref, would you please explian that? – Charmaine N Ndolo Jun 27 '17 at 6:05
• @CharmaineNNdolo I’ve got a detailed explanation here. – amd Jun 27 '17 at 6:46