Mathematics Asked by Future Math person on February 18, 2021

I’m asked if the matrices

$begin{pmatrix}

1 & 1\

1 & -1

end{pmatrix}$,

$begin{pmatrix}

0 & 2\

3 & 0

end{pmatrix}$, $begin{pmatrix}

2 & 1\

-3 & -2

end{pmatrix}$, $begin{pmatrix}

-1 & 4\

5 & 1

end{pmatrix}$

span $M_2 (mathbb{R})$.

I know that the traces of each of them are $0$ so they can’t possibly span $M_2 (mathbb{R})$ since you can’t write them as a linear combination with matrices that have non-zero traces.

However, I also attempted to do this with a system of equations. If I make a coefficient matrix from this, I get:

$begin{pmatrix}

1 & 0 & 2 & -1 \

1 & 2 & 1 & 4 \

1 & 3 & -3 & 5 \

-1 & 0 & -2 & 1 \

end{pmatrix}$.

After row reducing, I get:

$begin{pmatrix}

1 & 0 & 0 & -13/7 \

0& 1 & 0& 19/7 \

0 & 0 & 1 & 3/7 \

0 & 0 & 0 & 0 \

end{pmatrix}$

this means the last variable is free and I have a consistent solution and hence, should span $M_2 (mathbb{R})$ but it doesn’t.

Why is my reasoning wrong?

As the last row is equal to zero, you won't be able to generate a vector having the last coordinate not equal to zero.

Hence those four matrices can't span $M_2(mathbb R)$.

Correct answer by mathcounterexamples.net on February 18, 2021

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