Author Topic: Linear Algebra Questions (Read 902 times) Tweet Share

qshyrn · « **on:** March 14, 2011, 11:17:51 pm »

May need help on some questions from time to time.

Find a sqr (real) matrix $A$ that satisfies $AA^{T}=-I$

Mao · « **Reply #1 on:** March 15, 2011, 12:03:29 am »

I have a hunch this identity is false always. Let's have a look:

Firstly, let's express A as a bunch of vectors in $R^N$ , that is $A = [v_1, v_2,\cdots , v_N]$ , where v1, v2, ... vN are column vectors

Then, $A^T = \left[ \begin{array}{c} v_1 \\ v_2 \\ \vdots \\ v_N \end{array} \right]$ , where v1,v2,...,vN are row vectors

Let $A A^T = -I$ , this then implies $A^T A = (-I)^T = -I = A A^T$

So, we compute $A^T A = \left[ \begin{array}{c} v_1 \\ v_2 \\ \vdots \\ v_N \end{array}\right] [v_1,v_2,\cdots, v_N] = \left[ \begin{array}{cccc} v_1\cdot v_1 & v_1\cdot v_2 & \cdots & v_1 \cdot v_N \\<br /> v_2\cdot v_1 & v_2\cdot v_2 & \cdots & v_2 \cdot v_N \\<br /> \vdots & \vdots & \ddots & \vdots \\<br /> v_N\cdot v_1 & v_N\cdot v_2 & \cdots & v_N\cdot v_N \end{array} \right]$

At this point, all diagonal entries are required to be -1, that is $|v_1|^2=|v_2|^2=\cdots =|v_N|^2=-1$ , which is impossible.

So $A A^T = -I$ does not hold for real square matrices. (I'm fairly sure, have I made a big stupid somewhere?)

qshyrn · « **Reply #2 on:** March 16, 2011, 06:48:21 pm »

Quote from: qshyrn on March 14, 2011, 11:17:51 pm

May need help on some questions from time to time.
Find a sqr (real) matrix $A$ that satisfies $AA^{T}=-I$

thanks mao.think i agree with you.

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