Question

Recall that for a matrix A∈Rn×n,A is symmetric positive definite (SPD) if A=AT and xTAx>0 for all nonzero vectors x∈Rn. Use only this definition to prove the following properties of the SPD matrix A : (c) the submatrix A(1:k,1:k) is SPD for all k.

Answer 1

To prove the submatrix A(1:k, 1:k) of an SPD matrix A is also SPD, we show that the submatrix is symmetric and that it satisfies the positive definiteness condition using an appropriate extension of a nonzero vector from R^k to R^n.

Step-by-step explanation:

To prove that for a symmetric positive definite (SPD) matrix A the submatrix A(1:k, 1:k) is also SPD, we assume A is an n × n SPD matrix, meaning A = A^T and x^TAx > 0 for all nonzero vectors x ∈ R^n. Now we need to verify that the submatrix A(1:k, 1:k), which we'll denote as B, satisfies these two conditions for symmetry and positive definiteness.

Firstly, for symmetry, if A = A^T, this implies that B is also symmetric because the corresponding elements in the top-left k × k block of the larger symmetric matrix A are equal to their transposes; that is B = B^T.

Secondly, to verify positive definiteness, take any nonzero vector y ∈ R^k. We can extend y to a vector x ∈ R^n by appending n-k zeros: x = [y; 0; ... ; 0]. Because A is SPD, we know that x^TAx > 0 for all x ≠ 0. Notice that x^TAx = y^TBy since the rest of the entries in x are zeros and thus will not contribute to the sum. This means y^TBy > 0 for all nonzero y, which means that B is positive definite.