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Linear Algebra With Applications
Found in: Page 1
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

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Short Answer

Recall that a real square matrix A is called skew symmetric if AT=-A .

a. If A is skew symmetric, is A2 skew symmetric as well? Or is A2 symmetric?

b. If is skew symmetric, what can you say about the definiteness of A2 ? What about the eigenvalues of A2?

c. What can you say about the complex eigenvalues of a skew-symmetric matrix? Which skew-symmetric matrices are diagonalizable over ?

Therefore the solution is

a. A2 is symmetric.

b. A2 is a negative definite.

c. The zero matrix is the only skew-symmetric matrix that is diagonalizable over

See the step by step solution

Step by Step Solution

 Step 1: Determine the transpose of a symmetric matrix is the same.

a.) A symmetric matrix is equal to its transpose


and a skew symmetric matrix is a matrix whose transpose is equal to its negative


Now, if is a skew symmetric matrix, then as we just mentioned: AT=-A, now to check if A2 would be skew symmetric as well, we have:

A2=AAT =ATAT =-A-A =A2

therefore, A2 is not skew symmetric, it is just symmetric.

Step 2: Determine the certainty of A2

b.) We have


from part (a) we mentioned A that is said to be a skew matrix if AT=-A, so


therefore, for all x.qx0 , which implies that A2 is a negative definite (its eigenvalues are less than or equal to 0).

Step 3: Determine the skew symmetric matric

c.) Let be an eigenvector corresponding to the eigenvalue,


multiply both sides by v-T


note: the bar" -" above v is for complex conjugation.


for the left-hand side suppose we have column vectors of the same size, a and b, then aTb is a 1×1 matrix which we can think of as a scalar. Now, taking transposes, since aTb is 1×1, it is its own transpose, so


here let a=v and localid="1659623457872" b=Av , then for our left-hand side


since A is skew-symmetric, we have AT=-A . Substituting,


taking the conjugate of Av=λv yields Av¯=λv¯ , notice we can substitute this into the left-hand side above


Let λ=a+ib,

-a+ib=a+ib which implies a=0 and λ=bi. We also conclude that the zero matrix is the only skew-symmetric matrix that is diagonalizable over .

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