Finding SPD Matrix: Vanderberghe and Boyd Reader 6.4 Solution

[earlh]

Hi -- so, I'm working through a text on my own 5 years out of college having completely forgotten more or less all linear algebra. Here's my problem -- 6.4 from a reader by Vanderberghe and Boyd which is btw excellent.

Homework Equations

I have to show when a matrix is SPD. Setup:
let A which is n by n be SPD. For what values of scalar a is the matrix
B = \left[ \begin{array}{cc}<br /> A &amp; ae_{1} \\<br /> ae_{1}^{T} &amp; 1 \\<br /> \end{array} \right]
where e_{1} is the first vector in the usual basis

SPD? I need to find an interval for a so that B is SPD

The Attempt at a Solution

So I let v = \left[ \begin{array}{c} x \\ y \end{array} \right]
where x is (n, 1) and y is 1 by 1.

Multiplying things out, you get
v^{T} B v = [ x^{t} y ] \left[ \begin{array}{cc}<br /> A &amp; e_{1} \\<br /> e_{1}^{T} &amp; a \\<br /> \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right]

... = x^{T} A x + y^{2} + 2ay x^{T}e_{1} = x^{T}Ax + y^{2} + 2ayx_{1} where x_{1} is the first term in x
Obviously the first 2 terms are greater than or equal to zero, but I'm having trouble figuring out what a I use so that v can be arbitrary...Thanks in advance for any help

 

[vela]

Just to satisfy my curiosity, what does SPD stand for?

 

[earlh]

Just to satisfy my curiosity, what does SPD stand for?

Symmetric positive definite -- sorry for using jargon