# Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 35 exercise 2.15

Author: Panagiotis
25
Jun

In

[1, p. 35 exercise 2.15] we are asked to verify the formulas given for the gradient of a quadratic and linear form

[1, p. 31 (2.61)].
The corresponding formulas are

and

where

is a symmetric

matrix with elements

and

is a real

vector with elements

and

denotes the gradient of a real function in respect to

.

**Solution:**
Let

be the scalar function of the vector variable

defined by:

In order to identify the factors to the

vector component of

,

and its powers, we can rewrite the previous equation as:

The first derivative of (

3) in respect to

is given by:

and replacing the dummy indexes

and

of the two sums by

we obtain by considering the fact that the matrix is symmetric

the following relation:

Let

be the unit vector along the component

then the gradient of the matrix

can be written as:

which proves the first of the formulas

[1, p. 31 (2.61)].
In order to prove the second formula we proceed in a similar way by defining the scalar function

Thus the derivative in respect to

is simply:

and the gradient is given by

The last equation proves the second relation of the formulas

[1, p. 31 (2.61)]. QED.

[1] Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”, Prentice Hall, ISBN: 0-13-598582-X.
## One Response for "Steven M. Kay: “Modern Spectral Estimation – Theory and Applications”,p. 35 exercise 2.15"

[...] of the gradient of a quadratic form of a real matrix which is not symmetric. We reproduce from [2, relation (4) from the solution of exercise 2.15] the formula for the real matrix and the real vector [...]

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