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Find the dimension of the space of all symmetric nxn matrices.

Short Answer

Expert verified

The dimension of a nxn symmetric matrices is n2+n2which is spanned by

Span10…000…0⋮⋮⋱⋮00…0,10…000…0⋮⋮⋱⋮00…0,...,0…00⋮⋱0000010010,0…00⋮⋱0000000001.

Step by step solution

01

Determine the basis.

Consider the matrix A=a11a12…a1na12a22…a2n⋮⋮⋱⋮a1na2n…2nnwhere all aijare real.

The matrix A is symmetric if AT=Aand the general form of any skew-symmetric matrix is role="math" localid="1660131384364" [a11a12…a1na12a22…a2n⋮⋮⋱⋮a1na2n…2nn].

The total number of elements in the upper triangular matrix is n2+n2.

Simplify the equation role="math" localid="1660131413444" A=[a11a12…a1na12a22…a2n⋮⋮⋱⋮a1na2n…2nn]as follows.

role="math" localid="1660132125768" A=[a11a12…a1na12a22…a2n⋮⋮⋱⋮a1na2n…2nn]A=a110…000…0⋮⋮⋱⋮00…0+0a12…0a120…0⋮⋮⋱⋮00…0+...+0…00⋮⋱00000a(n-1)n00a(n-1)n0+0…00⋮⋱00000ann0000A=a1110…000…0⋮⋮⋱⋮00…0+a1201…010…0⋮⋮⋱⋮00…0+...an-1n0…00⋮⋱0000010010+ann0…00⋮⋱0000010001where10…000…0⋮⋮⋱⋮00…0,01…010…0⋮⋮⋱⋮00…0,...,0…00⋮⋱0000010010,0…00⋮⋱0000010001arelinearindependent.

.

Therefore, the matrix A is spanned by .

Span10…000…0⋮⋮⋱⋮00…0,01…010…0⋮⋮⋱⋮00…0,...,0…00⋮⋱0000010010,0…00⋮⋱0000010001

By the definition, the total number of elements in the setis .

10…000…0⋮⋮⋱⋮00…0,01…010…0⋮⋮⋱⋮00…0,...,0…00⋮⋱0000010010,0…00⋮⋱0000010001isn2+n2

Hence, the dimension of A is n2+n2which is spanned by

Span10…000…0⋮⋮⋱⋮00…0,01…010…0⋮⋮⋱⋮00…0,...,0…00⋮⋱0000010010,0…00⋮⋱0000010001.

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