Question

Hehe

Answer 1

Show that the 4 2x2 complex matrices below are the basis of a 4 dimensional real vector space.

\(\begin{pmatrix}
1 &0 \\
0 &1
\end{pmatrix} \), \(\begin{pmatrix}
0 &1 \\
1 &0
\end{pmatrix} \),\(\begin{pmatrix}
0 &-i \\
i &0
\end{pmatrix} \), and \(\begin{pmatrix}
1 &0 \\
0 &-1
\end{pmatrix} \)

Answer 2

Multiply the third matrix by i to get a real matrix. \(\mathbb{R}^4\) has four element in the basis. Write \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) as \(\begin{pmatrix}a\\c\\b\\d\end{pmatrix}\). Verify them as linearly independent and you are done.

Answer 3

You can't multiply a matrix by i, otherwise it would no longer be a real vector space.

Answer 4

boo! Matrices should be pure computation! just saying :/

Answer 5

Lol I don't even know what you mean

Answer 6

??? The fact that we need to do proofs in Linear Algebra is depressing. I rather just solve the matrices :)

Answer 7

But how could the four matrices be a basis for \(\mathbb{R}^4\) if there are i in the third matrix?

Answer 8

The matrices are the basis vectors :P

Answer 9

Yes but how? Any linear combination with the third matrix will result in a matrix with imaginary number in it. Clearly the new matrix could not be in \(\mathbb{R}^4\). If you discard the third matrix, you don't have enough matrices to form a basis in \(\mathbb{R}^4\).

Answer 10

The four matrices do not form a basis of \(\mathbb{R}^4\) but rather a vector space isomorphic to \(\mathbb{R}^4\). The isomorphism is multiplying the third basis matrix by i.

Answer 11

https://www.youtube.com/watch?v=BWM0RXg-uvI&feature=youtu.be&list=PLUl4u3cNGP60QlYNsy52fctVBOlk-4lYx&t=2674