# Constructing two symmetrical square matrices from a non-square matrix

Given two matrices:

$$A= \begin{pmatrix} 1 & 0 & 1 & 2\\ 2 & 1 & -1 & 1\\ 0 & 1 & 0 & 2\\ \end{pmatrix}$$

$$B= \begin{pmatrix} 2 & -1 & -2 \\ 3 & 4 & -3 \\ 0 & 1 & 2 \\ \end{pmatrix}$$

a) Construct two symmetric square matrices from matrix $A$
b) Calculate the symmetric and anti-symmetric part of matrix $B$

I don't know how to go about question a). I tried a LU Decomposition but it didn't give the desired outcome. Only other thing I could think of was to take the Lower matrix from the LU decomposition, then add it to its transpose, which of course gives a symmetric square matrix but it seemed too "forced" to me, plus it only gives one matrix instead of the two that are demanded. So I'm sure that there's another method that I'm not aware of.

As for question b) I don't understand what they mean by "symmetric and anti-symmetric parts" and what does it mean to calculate them.

Hope someone can help me solve this, or at least give me some hints. Thanks.

• question a) is really ambiguous, do you have more context about it? May 8 '18 at 14:52
• No I don't. That's all that was given in the question. What's one way you would go about it if you don't mind me asking? May 8 '18 at 15:10
• in real life? ask the professor May 8 '18 at 15:28
• @GuillermoMosse That would've been my go-to but it wasn't given by a professor I just stumbled upon it while preparing for an upcoming entry exam May 8 '18 at 15:56
• My guess for part (a) is that you’re simply meant to compute $A^TA$ and $AA^T$.
– amd
May 8 '18 at 17:59

For b): Prove that $B= \frac{1}{2}(B+B^t)+\frac{1}{2}(B-B^t)$