Exercise 4.2.22. For Z = < b m a t r i x

Exercise 4.2.22. For $Z = \begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ and $W = \begin{bmatrix} c & -d \\ d & c \end{bmatrix}$, determine $Z + W$ and $ZW$. For complex numbers $z = a + bi$ and $w = c + di$, determine $z + w$ and $zw$. What do you notice? Exercise 4.2.23. For $a \in \mathbb{C}$ the equation $a^2 = 1$ has exactly two solutions. Find eight matrices $A$ for which $A^2 = I_3$

Transcript

00:01 We're going to prove two parts.

00:04 First one called 4 .24 is that any metrics a can be written as the sum of a symmetric matrix and as q symmetric metrics.

00:18 In part 4 .25, we're going to prove that in part a that w intersection w2 is, sorry, sorry, intersection of w1 and w2 is the zero matrix, where where w is the vector space of square n times n matrices.

00:39 W is the subspace of symmetric matrices of n times n, and w is the subspace of q -symmetric matrices.

00:49 That is, the space or vector space of n times n matrices, is the direct sum of the subspace of symmetric matrices and the subspace of skewsymetric matrices.

01:07 And being the sum, a direct sum, means the intersection of the two subspaces is zero matrix.

01:17 So we're going to prove first 424.

01:23 And for that we are going to define s as a matrix 1⁄2 times a plus a transpose.

01:35 Now let's see this matrix has the same size as a, that is, if a is n times n, so will be s, because it's a sum to n times n matrices, a and a transpose, and that multiplied by a scalar has the same size n times n.

01:59 So we have that.

02:00 Now we can calculate s transpose, and there will be the transpose of one -half times a, plus a transpose, the scalar can be get out of these transposition.

02:18 And we get one half times the transpose of a plus a transpose.

02:23 Now we know the transpose of the sum of two matrices is the sum of the transposes of the matrices.

02:29 So we get one half times transpose of these first matrix a plus the transpose of these second matrix a transpose.

02:39 And we know if we transpose twice a matrix, we get the original matrix.

02:46 So we get one half times a transpose plus a.

02:51 The sum is commutatives of this one half times a plus a transpose.

02:57 But we can see easily here that this matrix is just s.

03:03 So this is equal to s.

03:05 So we have proved that s is equal to its transpose and that means that s is symmetrics is symmetric.

03:19 So we have that and now we define t equal one half times a minus a transpose so again because we have a difference of two matrices they must have the same size and that's the case because if a is n times n so will be a transpose so the difference is also n times n times the scalar is also n times n.

03:54 So t has the same size as a.

03:59 And now we calculate the transpose of t and we get, we know we can get up the scalar one half, so we get the one half times a minus a transpose transpose, and that is one half times transpose of the first matrix a minus transpose of the second metric.

04:25 A transpose and that is one half times a transpose minus transpose of the transpose of a is a and now we can get a negative out so we get negative one half times a minus a transpose because because we have to change signs of two matrices inside parentheses and we recognize here the matrix t again so we get negative in this case.

04:57 So we have proven that that t transpose is negative t.

05:07 That's just the definition of skew symmetric, so t is skew symmetric metrics.

05:20 But now we are going to prove s plus t is just one half times a plus a transpose, that's s over here plus t is one half times a minus a transpose so we get plus one half times a minus a transpose and that is one half is a common factor so we get a plus a transpose plus and a minus a transpose these two matrices cancel out they are the same with different signs so they that gives us a zero matrix and so that is the sum of these two terms here and so we get one -half times a plus a but that's two times matrix a so we get one half times two times metrics a and the coefficient one -half multiply by the coefficient 2 get coefficient 1 so you get a.

06:41 That is matrix a is equal to s plus t where s is symmetric matrix and t is skew symmetric.

07:05 So what we have proof here is that any matrix a, any square matrix a, can be written as a sum of a symmetric matrix plus skew symmetric matrix.

07:19 And it's clear that the two matrices, the symmetric one and the skew symmetric one, are constructed using only a because, for example, the symmetric matrix, one half times a plus a transpose uses only the information of matrix a, because we use a directly and it's transpose.

07:42 Same thing for the skew symmetry matrix...

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