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Chapter 8: Q42E (page 402)

Find the dimension of the spaceQn of all quadratic forms in n variables.

Short Answer

Expert verified

the solution is

鈭�dim(Qn)=nn+12

Step by step solution

01

 Find the dimension of the space Qn

The space of all quadratic formsqx in n variables is known as Qn.

Qn={q(x)=xTAx:

That is A, is an n-dimensional symmetric matrix}. As a result, the dimension of Qn={q(x)=xTAx:

The dimension of the symmetric matrix A of order n} that equals is the size of the symmetric matrix A of order n that equals nn+12.

SinceA=AT

鈭�dm(Qn)=n(n+1)2

02

conclusion

鈭�dim(Qn)=n(n+1)2

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Most popular questions from this chapter

The determinant of a negative definite4脳4matrix must be positive.

If Ais any symmetric 3x3matrix with eigenvalues -2,3, and 4, and u鈫�is a unit vector in鈩�3, what are the possible values of the dot productu鈫�.Au鈫�?

We say that ann脳nmatrix A is triangulizable ifis similar to an upper triangular n脳nmatrix B.

a. Give an example of a matrix with real entries that fails to be triangulizable over R.

b. Show that anyn脳nmatrix with complex entries is triangulizable over C . Hint: Give a proof by induction analogous to the proof of Theorem 8.1.1.

IfA is any symmetric2脳2matrix with eigenvalues -2 and 3, andu鈫�is a unit vector in鈩�2,鈩�2, what are the possible values of||Au炉||? Explain your answer geometrically, using Example 4as a guide.

Find the singular values of A=[1224]. Find a unit vectorv鈫�such that||Av1鈫�||=蟽1. Sketch the image of the unit circle.
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