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I am interested in verifying that my understanding of the basis for $3 \times 3$ (or even $n \times n$ matrices) that have a trace $= 0$ is along the right path.

$$ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{pmatrix} $$ With the above $3 \times 3$ matrix with trace $0$ know that

$a + e + i = 0$

$a = -e - i$

$e = -i -a$

$i = -a -e$

Outside of the elementary bases making up everything except the diagonal, I thought my last basis was

$$ \begin{pmatrix} -e-i & 0 & 0 \\ 0 & -i-a & 0 \\ 0 & 0 & -a-e \\ \end{pmatrix} $$

But by inspection, this does not seem to account for the instance where one of the diagonals is 0 and the other two are not.

My last consideration that I think is more accurate would be splitting the above matrix into the following combinations: $$ \begin{pmatrix} a & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & 0 \\ \end{pmatrix} $$ $$ \begin{pmatrix} 0 & 0 & 0 \\ 0 & -a & 0 \\ 0 & 0 & a \\ \end{pmatrix} $$

But this also doesn't feel quite correct. I feel I am close but missing one additional consideration.

I have also found other posts concerning the dimensions like here, but there was never a concrete basis provided to go off of.

Thanks!

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  • $\begingroup$ A basis for this space will be a set of eight matrices. I don't see anything like that in your question. $\endgroup$
    – Angina Seng
    Mar 15, 2018 at 1:50
  • $\begingroup$ What is unclear from the accepted answer from your linked post? $\endgroup$
    – B. Mehta
    Mar 15, 2018 at 1:51
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    $\begingroup$ I suggest you use \pmatrix in place of \matrix $\endgroup$
    – Angina Seng
    Mar 15, 2018 at 1:52
  • $\begingroup$ @LordSharktheUnknown The other elementary matrices that do not make up the diagonal I understand, it is the last matrices that complete the basis that account for values in the diagonal I am unsure of how to find. $\endgroup$
    – Bailey Hollingsworth
    Mar 15, 2018 at 1:52
  • $\begingroup$ Do you know how to find a basis for the nullspace of a general linear transformation? You can use the same process. $\endgroup$
    – Matthew Leingang
    Mar 15, 2018 at 1:54

2 Answers 2

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For your matrix, there is only a single constraint.

Your problem have $9$ variables and $1$ constraint.

$$a+e+i = 0$$

The nullity is $9-1=8$.

In general the matrix can be written as

\begin{bmatrix} -e-i & b & c \\ d & e & f \\ g & h & i \end{bmatrix}

Let me write out a few elements of the possible basis, by observing the places where $e$ appears, we can choose

$$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

By observing places where $i$ appears, we can choose

$$\begin{bmatrix} -1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{bmatrix}$$

By observing places where $b$ appears, we can choose

$$\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}$$

Try to find the other $5$ elements.

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  • $\begingroup$ When you observe places where i appears, would it not be -1 at (1,1) and 1 at (3,3)? $\endgroup$
    – Bailey Hollingsworth
    Mar 15, 2018 at 2:03
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    $\begingroup$ oops, thanks for pointing out the mistake. $\endgroup$
    – Siong Thye Goh
    Mar 15, 2018 at 2:03
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Your condition on trace makes the space to be an $8$ dimensional space. For a basis you may try $$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \\ \end{pmatrix}$$

and $$\begin{pmatrix} 0 & 0 & 0 \\ 0 & 1& 0 \\ 0 & 0 & -1 \\ \end{pmatrix}$$

and the other $6$ elementary basis matrices whose diagonal elements are $0$

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