2
$\begingroup$

Is the set of matrices $\{\begin{bmatrix}0&1\\0&1\end{bmatrix},\begin{bmatrix}1&1\\2&2\end{bmatrix},\begin{bmatrix}1&4\\2&2\end{bmatrix},\begin{bmatrix}5&4\\2&6\end{bmatrix}\}$ in $M_{22}$ linearly independent or linearly dependent?

For some reason I'm stuck.

What I have done:

$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = 0$

What I was thinking of doing is:

$A_1(det(A_1)) +A_2(det(A_2) + A_3(det(A_3)) + A_4(det(A_4)) = 0$

I don't know if that's right but I do know I'm on the right track! Any help would be much appreciated!

$\endgroup$
2
$\begingroup$

Looking into the determinant in this way is heading in the wrong direction. The determinant is not preserved under addition. To illustrate: $$ \overbrace{\begin{bmatrix} 2 & 2 \\ -1 & 2 \\ \end{bmatrix}}^{\text{det}=6} + \overbrace{\begin{bmatrix} 0 & -1 \\ 1 & -2 \\ \end{bmatrix}}^{\text{det}=1} = \overbrace{\begin{bmatrix} 2 & 1 \\ 0 & 0 \\ \end{bmatrix}}^{\text{det}=0}.$$

To continue, notice that the system of equations $$A_1\begin{bmatrix}0&1\\0&1\end{bmatrix} + A_2\begin{bmatrix}1&1\\2&2\end{bmatrix} + A_3\begin{bmatrix}1&4\\2&2\end{bmatrix} + A_4\begin{bmatrix}5&4\\2&6\end{bmatrix} = \begin{bmatrix}0&0\\0&0\end{bmatrix}$$ is equivalent to \begin{align*} A_2+A_3+5A_4 &= 0 \\ A_1+A_2+4A_3+4A_4 &= 0 \\ 2A_2+2A_3+2A_4 &= 0 \\ A_1+2A_2+2A_3+6A_4 &= 0. \end{align*} We can solve this in the usual way (i.e., Gaussian elimination).

Alternatively, we write the matrices as the rows of a $4 \times 4$ matrix $$ \begin{bmatrix} 0 & 1 & 0 & 1 \\ 1 & 1 & 2 & 2 \\ 1 & 4 & 2 & 2 \\ 5 & 4 & 2 & 6 \\ \end{bmatrix}. $$ If the rows of this matrix are linearly independent (which occurs if and only if the matrix has non-zero determinant), then the original set of matrices is linearly independent.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

Hint : You get 4 equations with 4 unknowns if you solve the matrix equation elementwise. Note, that the sum is the null-matrix, not the number 0.

You do not need determinants.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

Your first equation is correct, where $A_1,..A_4$ are scalars.

Note, that equation gives you:

$A_2+A_3+5A_4=0$

$A_1+A_2+4A_3+4A_4=0$

$2A_2+2A_3+3A_4=0$

and $A_1+2A_2+2A_3+6A_4=0$

Equations $1$ and $3$ then imply that $A_4=0$.

Equations $3$ and $4$ imply that $A_1=0$.

Ans then equations $1$ and $2$ imply that $A_2=A_3=0$.

So these are indeed linearly ind.

| cite | improve this answer | |
$\endgroup$

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.