Question

Show that if \(B\) is \(m \times n\), then \({B^T}B\) is positive semidefinite; and if \(B\) is \(n \times n\) and invertible, then \({B^T}B\) is positive definite.

Short answer

It is proved that if \(B\) is \(m \times n\), then \({B^T}B\) is positive semidefinite; and if \(B\) is \(n \times n\) and invertible, then \({B^T}B\) is positive definite.

Step-by-step solution

Step 1: Find the characteristic polynomial

We need to prove that\({B^T}B\)is a symmetric matrix and the quadratic form\({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} \ge 0\)for all\({\bf{x}}\)that proves\({B^T}P\)positive semi-definite.

Now consider,

\(\begin{aligned}{}{\left( {{B^T}B} \right)^T} &= {B^T}{\left( {{B^T}} \right)^T}\\ &= {B^T}B\end{aligned}\)

Thus, the matrix \({B^T}B\) is a symmetric matrix.

Show that \({B^T}B\) is positive semi definite

Consider the quadratic form \({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}}\).

\(\begin{aligned}{}{{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} &= \left( {{{\bf{x}}^T}{B^T}} \right)B{\bf{x}}\\ &= {\left( {B{\bf{x}}} \right)^T}B{\bf{x}}\\ &= {\left\| {B{\bf{x}}} \right\|^2}\end{aligned}\)

Thus, \({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} \ge 0\).

Therefore, the matrix \({B^T}B\) is positive semi-definite.

Show that \({B^T}B\) is positive definite

Assume \({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} = 0\)then we have,

\(\begin{aligned}{}{{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} &= 0\\\left( {{{\bf{x}}^T}{B^T}} \right)B{\bf{x}} &= 0\\{\left( {B{\bf{x}}} \right)^T}B{\bf{x}} &= 0\\{\left\| {B{\bf{x}}} \right\|^2} &= 0\end{aligned}\)

It is given that the matrix\(B\)is invertible, hence the equation\(B{\bf{x}} = 0\)has a trivial solution only. Therefore, we have,\({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} = 0\)then\({\bf{x}} = 0\)and\({\bf{x}} \ne 0\),\({{\bf{x}}^T}\left( {{B^T}B} \right){\bf{x}} > 0\).

Thus,\({B^T}B\)is a positive definite.

It is proved that if \(B\) is \(m \times n\), then \({B^T}B\)is positive semidefinite; and if\(B\) is \(n \times n\) and invertible, then \({B^T}B\)is positive definite.