How to Find the Basis for a Specific Vector Space?

Question

I have the following homework problem:

In Exercises 1-4, $W$ is a subspace of the vector space $V$ of all $(2 \times 2)$ matrices. A matrix $A$ in $W$ is written as A = $$ \begin{bmatrix} a & b \\ c & d \\ \end{bmatrix} $$ In each case exhibit a basis for $W$.

1. $\quad W = \{A: a+ b + c + d = O\}$
2. $\quad W = \{A: a= -d, b = 2d, c = -3d\}$
3. $\quad W = \{A: a = 0\}$
4. $\quad W = \{A: b = a - c, d = 2a + c\}$

I'm having trouble understanding number four, and I can't tell if I'm just overthinking it, or if there is some more depth to it. So far, all I've been doing is guess and check; but is it just as simple as putting a '$1$' in each position of the $(2 \times 2)$ matrix surrounded by zeroes?

Answer 1

That the elements are written as matrices is a bit of red herring. We never use matrix multiplication, so it might be easier to think of the matrix $\begin{bmatrix}a&b \\ c&d\end{bmatrix}$ as the column vector $$\begin{bmatrix}a\\ b\\c\\d\end{bmatrix}.$$

Let’s consider part 4. As $W$ is given by equations, we want to calculate a kernel (or null-space). Namely, we can rewrite the given equations as $$ a - b - c = 0 $$ and $$ 2a + c - d = 0. $$ We can write this as a matrix and perform Gaussian elimination: $$ \begin{bmatrix}1&-1&-1&0\\ 2&0&1&-1\end{bmatrix} \leadsto \begin{bmatrix}1&0&1/2&-1/2\\ 0&1&3/2&-1/2\end{bmatrix}. $$ A basis of the null-space is $$ \begin{bmatrix}-1/2\\-3/2\\1\\0\end{bmatrix},\begin{bmatrix}1/2\\1/2\\0\\1\end{bmatrix}, $$ or in matrix form again $$ \begin{bmatrix}-1/2&-3/2\\1&0\end{bmatrix},\begin{bmatrix}1/2&1/2\\0&1\end{bmatrix}. $$ (You can multiply these by $2$ to get integer entries, if you want.)

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As an aside, the switch from matrices to column vectors corresponds to representing matrices in the basis $$ \begin{bmatrix}1&0\\0&0\end{bmatrix}, \begin{bmatrix}0&1\\0&0\end{bmatrix}, \begin{bmatrix}0&0\\1&0\end{bmatrix}, \begin{bmatrix}0&0\\0&1\end{bmatrix}. $$ Calling these $E_1, \dots, E_4$ to shorten the notation a bit, we can write any matrix in the form $$ \begin{bmatrix}a&b\\c&d\end{bmatrix} = a E_1 + b E_2 + c E_3 + d E_4, $$ so the column vector above is just the coordinate vector of the matrix with respect to this basis.

In your case, the equations are already given in terms of the coordinates (i.e. the given equations use $a, b, c, d$) but choosing a basis like this would help if you had equations like $A = 2A - I$.

Answer 2

For number four, it is equivalent to finding a basis for the image of the matrix $$A=\begin{bmatrix} 1&&0&&0&&0\\ 1&&0&&-1&&0\\ 0&&0&&1&&0\\ 2&&0&&1&&0 \end{bmatrix}$$

where $A$ acts on $\mathbb{R}^4$ because if $$\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$ is in the image of $A$, then there is some vector $$\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}$$ such that $$A\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix}=\begin{bmatrix}x_1\\x_1-x_3\\x_3\\2x_1+x_3\end{bmatrix}=\begin{bmatrix}a\\b\\c\\d\end{bmatrix}$$ which implies \begin{align} a&=x_1\\ c&=x_3\\ b&=a-c\\ d&=2a+c \end{align} One such basis is given by $$\left\{\begin{bmatrix}1&&1\\0&&2\end{bmatrix},\begin{bmatrix}0&&-1\\1&&1\end{bmatrix}\right\}$$ which is just the non-zero columns of $A$.