Comprehensive Exercises on Quadratic Forms

Approach: Build the symmetric matrix of the quadratic form by the standard rule: diagonal entries come from the squared terms and each off-diagonal entry is half of the mixed-term coefficient.

Steps:

1. Diagonal entries: $a_{11} = 2$, $a_{22} = 3$
2. Off-diagonal entries: $a_{12} = a_{21} = \frac{4}{2} = 2$
3. Matrix representation: $A = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix}$

Answer:$A = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix}$

Approach: Diagonalise the symmetric matrix of the form, normalise the eigenvectors into an orthogonal matrix, and read the standard form off the eigenvalues.

Steps:

1. Matrix representation: $A = \begin{pmatrix} 2 & 2 \\ 2 & 3 \end{pmatrix}$
2. Characteristic equation: $|A - \lambda I| = \lambda^2 - 5\lambda + 2 = 0$
3. Eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 4$
4. Eigenvectors:
   • $\lambda_1 = 1$: $\vec{x}_1 = (1, -1)$
   • $\lambda_2 = 4$: $\vec{x}_2 = (1, 1)$
5. Normalisation:
   • $\vec{q}_1 = \frac{1}{\sqrt{2}}(1, -1)$
   • $\vec{q}_2 = \frac{1}{\sqrt{2}}(1, 1)$
6. Orthogonal matrix: $Q = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\ -1 & 1 \end{pmatrix}$
7. Standard form: $Q = y_1^2 + 4y_2^2$

Answer:$Q = y_1^2 + 4y_2^2$

Approach: Evaluate the form on an arbitrary nonzero vector. If the value is positive for every such vector, the form is positive definite by definition.

Steps:

1. Matrix representation: $A = \begin{pmatrix} 1 & 0 \\ 0 & 2 \end{pmatrix}$
2. For any nonzero vector $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, $Q(x) = x_1^2 + 2x_2^2 > 0$
3. Hence the form is positive definite

Answer:It is a positive definite quadratic form

Approach: Recall the statement of the law of inertia, which connects the congruence of symmetric matrices with the invariance of their signature.

Steps:

The law of inertia: let $A$ be a real symmetric matrix of order $n$. Then the positive index of inertia and the negative index of inertia of $A$ are unchanged under congruence transformations.

Answer:Congruence transformations do not change the positive or negative indices of inertia

Approach: Try the eigenvalue criterion and the leading principal minor criterion; each of them decides the question independently, so agreeing results confirm the computation.

Steps:

1. Method one: compute the eigenvalues
   • Characteristic equation: $|A - \lambda I| = \lambda^2 - 4\lambda + 3 = 0$
   • Eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 3$, both positive
2. Method two: check the leading principal minors
   • $|A_1| = 2 > 0$
   • $|A_2| = 3 > 0$

Answer:It is a positive definite matrix

Approach: Collect the coefficients of the quadratic form into a symmetric matrix, halving the coefficients of the mixed terms for the off-diagonal entries.

Steps:

1. Diagonal entries: $a_{11} = 1$, $a_{22} = 2$, $a_{33} = 3$
2. Off-diagonal entries: $a_{12} = a_{21} = \frac{2}{2} = 1$, $a_{13} = a_{31} = \frac{2}{2} = 1$, $a_{23} = a_{32} = \frac{4}{2} = 2$
3. Matrix representation: $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{pmatrix}$

Answer:$A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{pmatrix}$

Approach: Absorb the cross term into a perfect square and then rename the new combination as a variable, which removes the mixed term entirely.

Steps:

1. $Q(x_1, x_2) = x_1^2 + 4x_1x_2 + 4x_2^2$
2. $= (x_1 + 2x_2)^2$
3. Put $y_1 = x_1 + 2x_2$, $y_2 = x_2$
4. Then $Q = y_1^2$

Answer:$Q = y_1^2$, where $y_1 = x_1 + 2x_2$, $y_2 = x_2$

Approach: Evaluate the form for an arbitrary nonzero vector. A value that is always negative shows that the form cannot be positive definite, but is negative definite.

Steps:

1. Matrix representation: $A = \begin{pmatrix} -1 & 0 \\ 0 & -2 \end{pmatrix}$
2. For any nonzero vector $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, $Q(x) = -x_1^2 - 2x_2^2 < 0$
3. Hence the form is negative definite

Answer:It is a negative definite quadratic form

Approach: Diagonalise the symmetric matrix of the form, normalise the eigenvectors so that they form an orthogonal basis, and then read the standard form off the eigenvalues.

Steps:

1. Matrix representation: $A = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{pmatrix}$
2. Characteristic equation: $|A - \lambda I| = -\lambda^3 + 6\lambda^2 - 11\lambda + 6 = 0$
3. Eigenvalues: $\lambda_1 = 1$, $\lambda_2 = 2$, $\lambda_3 = 3$
4. Eigenvectors:
   • $\lambda_1 = 1$: $\vec{x}_1 = (1, -1, 0)$
   • $\lambda_2 = 2$: $\vec{x}_2 = (1, 0, -1)$
   • $\lambda_3 = 3$: $\vec{x}_3 = (1, 1, 1)$
5. Normalisation:
   • $\vec{q}_1 = \frac{1}{\sqrt{2}}(1, -1, 0)$
   • $\vec{q}_2 = \frac{1}{\sqrt{2}}(1, 0, -1)$
   • $\vec{q}_3 = \frac{1}{\sqrt{3}}(1, 1, 1)$
6. Standard form: $Q = y_1^2 + 2y_2^2 + 3y_3^2$

Answer:$Q = y_1^2 + 2y_2^2 + 3y_3^2$

Approach: Compute the eigenvalues. A zero eigenvalue blocks positive definiteness, while the absence of negative eigenvalues still allows a semidefinite verdict.

Steps:

1. Characteristic equation: $|A - \lambda I| = \lambda^2 - 2\lambda = 0$
2. Eigenvalues: $\lambda_1 = 0$, $\lambda_2 = 2$
3. There is a zero eigenvalue, so the matrix is not positive definite
4. But all eigenvalues are nonnegative, so it is positive semidefinite

Answer:It is a positive semidefinite matrix

Approach: Combine the eigenvalue criterion with the fact that the determinant of a matrix is the product of its eigenvalues.

Steps:

1. Let $A$ be a positive definite matrix
2. All eigenvalues of a positive definite matrix are positive
3. The determinant is the product of the eigenvalues: $|A| = \prod_{i=1}^n \lambda_i$
4. All eigenvalues are positive, so their product is positive as well
5. Therefore $|A| > 0$

Answer:The proof is complete

Approach: Group the terms so that the cross terms are absorbed into perfect squares one after another, then introduce new variables for the completed expressions.

Steps:

1. $Q = x_1^2 + 2x_1x_2 + 2x_1x_3 + 2x_2^2 + 4x_2x_3 + 3x_3^2$
2. $= (x_1 + x_2 + x_3)^2 - x_2^2 - x_3^2 - 2x_2x_3 + 2x_2^2 + 4x_2x_3 + 3x_3^2$
3. $= (x_1 + x_2 + x_3)^2 + x_2^2 + 2x_2x_3 + 2x_3^2$
4. $= (x_1 + x_2 + x_3)^2 + (x_2 + x_3)^2 + x_3^2$
5. Put $y_1 = x_1 + x_2 + x_3$, $y_2 = x_2 + x_3$, $y_3 = x_3$
6. Then $Q = y_1^2 + y_2^2 + y_3^2$

Answer:$Q = y_1^2 + y_2^2 + y_3^2$, where $y_1 = x_1 + x_2 + x_3$, $y_2 = x_2 + x_3$, $y_3 = x_3$

Approach: Look for two test vectors that make the form positive and negative respectively; a form that takes both signs is indefinite.

Steps:

1. Matrix representation: $A = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$
2. For the vector $x = (1, 0)$, $Q(x) = 1 > 0$
3. For the vector $x = (0, 1)$, $Q(x) = -1 < 0$
4. The form takes both positive and negative values, so it is indefinite

Answer:It is an indefinite quadratic form

Approach: Combine the definition of congruence with the definition of positive definiteness: substitute the congruent matrix into the quadratic form and use the invertibility of the change of basis.

Steps:

1. Let $A$ be a positive definite matrix and let $B = P^TAP$ be a congruence transformation
2. For any nonzero vector $y$, $y^TBy = y^TP^TAPy = (Py)^TA(Py)$
3. Since $P$ is an invertible matrix, $Py$ is also a nonzero vector
4. Since $A$ is positive definite, $(Py)^TA(Py) > 0$
5. Therefore $y^TBy > 0$, that is, $B$ is positive definite as well

Answer:The proof is complete

Approach: Apply the eigenvalue criterion and the leading principal minor criterion, and compare the two computations before concluding.

Steps:

1. Method one: compute the eigenvalues
   • Characteristic equation: $|A - \lambda I| = (3-\lambda)^2 - 1 = \lambda^2 - 6\lambda + 8 = 0$
   • Eigenvalues: $\lambda_1 = 2$, $\lambda_2 = 4$, both positive
2. Method two: check the leading principal minors
   • $|A_1| = 3 > 0$
   • $|A_2| = 8 > 0$

Answer:It is a positive definite matrix