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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