Review

May 31, 2019

1. Do the invertible matrices form a subspace of the vector space by all $5\times 5$ matrices?

   First remember for a matrix to be invertible, it has to be a square matrix. After lec. 8, in fact an invertible matrix must be in full rank. In full rank means a square matrix has no dependency in rows or columns. Back to the question, the answer is no. Let’s use a $2\times 2$ example to illustrate:

   $$A=\begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix},B=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix},A+B=\begin{bmatrix} 1 & 1\\ 1 & 1 \end{bmatrix}$$

   Both $A$ and $B$ are $2\times 2$ invertible matrices but their sum is not. We can form similar cases in $5\times 5$ matrices. Inversely, do the singular matrices form a subspace of the vector space by all $5\times 5$ matrices? The answer is no again:

   $$A=\begin{bmatrix} -1 & 0\\ 0 & 0 \end{bmatrix},B=\begin{bmatrix} 1 & 1\\ 1 & 0 \end{bmatrix},A+B=\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}$$

2. Find the basis of null space for the following matrix $B$ without multiplying the matrix out

   $$B=CD=\begin{bmatrix} 1 & 1 & 0\\ 0 & 1 & 0\\ 1 & 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & -1 & 2\\ 0 & 1 & 1 & -1\\ 0 & 0 & 0 & 0 \end{bmatrix}$$

   In fact it’s not hard to multiply these two matrices, from $C$ we see the answer is just one row adding another. But one finding is important. $C$ is an invertible matrix. So it has $C^{-1}C=0$. Then we have the following:

   $$\begin{align} B\mathbf x&=0\\ CD\mathbf x&=\mathbf 0\\ C^{-1} CD\mathbf x&=C^{-1}\mathbf 0\\ D\mathbf x&=\mathbf 0 \end{align}$$

   That is, $N(B)=N(D)$. The nullspace of $C$ does not affect the nullspace of $D$ at all. And $D$ is already in rref, the third and fourth column is just free variables, so the basis is

   $$\begin{bmatrix} 1\\ -1\\ 1\\ 0\\ \end{bmatrix} ,\begin{bmatrix} -2\\ 1\\ 0\\ 1\\ \end{bmatrix}$$

3. True or False: $A$ and $-A$ share the same four subspaces:

   Yes. A subspace always carry a constant in front of it like $c\begin{bmatrix} 1\\ 1\\ -1\\ \end{bmatrix}$ to extend the basis to be a space. If $\mathbf v$ is in the subspace, so does $-\mathbf v$.

4. True or False: If $A$ and $B$ share the same four subspaces, is it true that $A=cB$.

   No. An example will be

   $$A=\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix},B=\begin{bmatrix} 1 & 0 & 0\\ 0 & 2 & 0\\ 0 & 0 & 1 \end{bmatrix}$$

   They are both invertible matrices so the nullspace and left nullspace are just zero. And they share the column and row space. One open question you might ask is if $A,B$ have same four subspaces, what property do they? (I don’t know the anwer)