"Linear Algebra" — Practice

Question 1

A 4 by 6 matrix has rank 3. What is the dimension of its null space?

Accepted Answer

C

Question 2

If A is a 3 by 3 matrix with det

Accepted Answer

C

Question 3

The eigenvalues of a 3 by 3 matrix A are 1, 2 and 4. What is the determinant of A inverse?

Accepted Answer

B

Question 4

For what values of a does the system x + y + z = 1, x + 2y + 4z = a, x + 4y + 10z = a squared have a solution?

Accepted Answer

Row-reduce the augmented matrix, comparing rank(A) with rank of the augmented matrix.

Subtract row 1 from row 2: y + 3z = a minus 1.
Subtract row 1 from row 3: 3y + 9z = a squared minus 1.

Notice that the third row's left side is exactly 3 times the second's, so subtracting 3 times the new row 2 from the new row 3 gives 0 = a squared minus 1 minus 3(a minus 1).

Simplify the right side: a squared minus 3a plus 2, which factors as (a - 1)(a - 2).

The left side is zero, so consistency requires (a - 1)(a - 2) = 0, giving a = 1 or a = 2.

For any other value of a, the reduced system contains a zero row on the left with a non-zero entry on the right, so rank(A) = 2 while the augmented rank is 3 and the system is inconsistent.

At a = 1 or a = 2 both ranks are 2, which is less than the 3 unknowns, so there are infinitely many solutions with 3 minus 2 = 1 free parameter.

Question 5

A 2 by 2 matrix has trace 7 and determinant 12. Find its eigenvalues, and then the eigenvalues of A squared minus 3A.

Accepted Answer

The eigenvalues satisfy sum = 7 and product = 12, so they are roots of lambda squared minus 7 lambda plus 12 = 0.

Factoring gives (lambda - 3)(lambda - 4) = 0, so the eigenvalues are 3 and 4.

For a polynomial in A, the eigenvalues are that same polynomial applied to each eigenvalue, and the eigenvectors are unchanged.

For lambda = 3: 9 minus 9 = 0.
For lambda = 4: 16 minus 12 = 4.

So A squared minus 3A has eigenvalues 0 and 4.

Note the consequence: one eigenvalue is 0, so A squared minus 3A is singular. That is worth reading off directly, because a follow-up part often asks for its determinant, which is 0 times 4 = 0.

Question 6

Determine whether the matrix with rows (5, 4) and (0, 5) is diagonalisable.

Accepted Answer

The matrix is upper triangular, so its eigenvalues are the diagonal entries: 5 and 5.

The eigenvalue 5 therefore has algebraic multiplicity 2.

For the geometric multiplicity, form A minus 5I, which has rows (0, 4) and (0, 0). Its rank is 1, since one non-zero row remains.

Geometric multiplicity = n minus rank = 2 minus 1 = 1.

Since 1 is strictly less than the algebraic multiplicity 2, the matrix has only one independent eigenvector and is not diagonalisable.

Contrast this with the matrix 5I, which also has the repeated eigenvalue 5 but whose A minus 5I is the zero matrix of rank 0, giving geometric multiplicity 2 and a diagonalisable — indeed already diagonal — matrix. A repeated eigenvalue alone never settles the question; the rank computation does.

Question 7

For A with rows (1, 2) and (3, 4), use Cayley-Hamilton to express A inverse in terms of A and I, and hence compute it.

Accepted Answer

The trace is 1 plus 4 = 5 and the determinant is 4 minus 6 = -2.

The characteristic equation is lambda squared minus 5 lambda minus 2 = 0.

By Cayley-Hamilton, A squared minus 5A minus 2I = 0.

Rearrange to isolate the identity: 2I = A squared minus 5A, so I = (A squared minus 5A)/2.

Now multiply the original relation by A inverse, which exists because the determinant is non-zero: A minus 5I minus 2A inverse = 0.

So A inverse = (A minus 5I)/2.

Compute it: A minus 5I has rows (1 - 5, 2) = (-4, 2) and (3, 4 - 5) = (3, -1). Dividing by 2 gives rows (-2, 1) and (1.5, -0.5).

Verify against the standard 2 by 2 formula, which swaps the diagonal, negates the off-diagonal and divides by the determinant: rows (4, -2) and (-3, 1) divided by -2 gives (-2, 1) and (1.5, -0.5). The two agree.

The general result worth remembering is that Cayley-Hamilton always expresses A inverse as a polynomial in A of degree one less than the matrix order.

Question 8

Find the Doolittle LU decomposition of the matrix with rows (1, 2, 4), (3, 8, 14), (2, 6, 13), and state why it exists without pivoting.

Accepted Answer

Doolittle sets the diagonal of L to 1, so L has unknowns l21, l31, l32 below the diagonal and U is upper triangular.

Row 1 of the product gives the first row of U directly: u11 = 1, u12 = 2, u13 = 4.

Row 2: l21 times u11 = 3, so l21 = 3. Then l21 u12 + u22 = 8 gives 6 + u22 = 8, so u22 = 2. And l21 u13 + u23 = 14 gives 12 + u23 = 14, so u23 = 2.

Row 3: l31 times u11 = 2, so l31 = 2. Then l31 u12 + l32 u22 = 6 gives 4 + 2 l32 = 6, so l32 = 1. Finally l31 u13 + l32 u23 + u33 = 13 gives 8 + 2 + u33 = 13, so u33 = 3.

So L has rows (1,0,0), (3,1,0), (2,1,1) and U has rows (1,2,4), (0,2,2), (0,0,3).

The factorisation exists without pivoting because every leading principal minor is non-zero: the first is 1, the second is 8 minus 6 = 2, and the third is the determinant, which equals the product of the diagonal of U, namely 1 times 2 times 3 = 6.

That last identity is also the fastest check: since L has unit diagonal, det A must equal the product of U's diagonal entries.

Question 9

A is a real 3 by 3 matrix satisfying A squared = A and having rank 2. Determine its eigenvalues with multiplicities, its trace and its determinant.

Accepted Answer

A squared equals A makes A idempotent. If lambda is an eigenvalue with eigenvector X, then A squared X = lambda squared X and also A X = lambda X, so lambda squared = lambda, giving lambda = 0 or lambda = 1.

The multiplicities follow from the rank. An idempotent matrix is diagonalisable, because the polynomial x squared minus x has distinct roots, so the geometric multiplicity of each eigenvalue equals its algebraic multiplicity.

The eigenvalue 1 has multiplicity equal to the rank, since the image of an idempotent matrix is exactly its eigenspace for 1. With rank 2, the eigenvalue 1 has multiplicity 2.

The remaining eigenvalue is 0 with multiplicity 3 minus 2 = 1, which matches the nullity.

Trace = sum of eigenvalues = 1 + 1 + 0 = 2, which equals the rank. This is the general fact that the trace of an idempotent matrix equals its rank.

Determinant = product of eigenvalues = 1 times 1 times 0 = 0, so A is singular, as it must be since its rank 2 is less than 3.

Practice — "Linear Algebra"

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