Kainui
  • Kainui
Hehe
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Kainui
  • Kainui
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 \(\begin{pmatrix} 1 &0 \\ 0 &-1 \end{pmatrix} \)
thomas5267
  • thomas5267
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.
Kainui
  • Kainui
You can't multiply a matrix by i, otherwise it would no longer be a real vector space.

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UsukiDoll
  • UsukiDoll
boo! Matrices should be pure computation! just saying :/
Kainui
  • Kainui
Lol I don't even know what you mean
UsukiDoll
  • UsukiDoll
??? The fact that we need to do proofs in Linear Algebra is depressing. I rather just solve the matrices :)
thomas5267
  • thomas5267
But how could the four matrices be a basis for \(\mathbb{R}^4\) if there are i in the third matrix?
Kainui
  • Kainui
The matrices are the basis vectors :P
thomas5267
  • thomas5267
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 \(\mathbb{R}^4\).
thomas5267
  • thomas5267
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.
Kainui
  • Kainui
https://www.youtube.com/watch?v=BWM0RXg-uvI&feature=youtu.be&list=PLUl4u3cNGP60QlYNsy52fctVBOlk-4lYx&t=2674

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