6.2 Eigenvalues, Eigenvectors & the Cayley-Hamilton Theorem

The Eigenvalue Problem

For a square matrix $A$ of order $n$, a non-zero vector $X$ and a scalar $\lambda$ satisfying

$AX = \lambda X \quad \Longrightarrow \quad (A - \lambda I)X = 0$

are called an eigenvector and its eigenvalue. Because $X \neq 0$ is demanded, the homogeneous system must be singular, which gives the characteristic equation

$\det(A - \lambda I) = 0$

This is exactly the condition developed in the previous section, and it is the same mathematics that produces buckling loads for a strut and natural frequencies for a shaft — a genuine cross-link inside Paper-II.

For a $3 \times 3$ matrix the characteristic polynomial expands to

$\lambda^3 - S_1\lambda^2 + S_2\lambda - S_3 = 0$

where $S_1$ is the trace (sum of diagonal entries), $S_2$ is the sum of the three principal $2\times2$ minors, and $S_3 = \det A$.

The Two Shortcuts That Win Marks

Because CIL gives one mark per item with no negative marking and a 3-hour budget for 200 questions, speed matters more than rigour. Two identities let you answer most eigenvalue items in seconds:

$\sum_{i=1}^{n} \lambda_i = \text{trace}(A), \qquad \prod_{i=1}^{n} \lambda_i = \det(A)$

If a question gives you two of three eigenvalues and the trace, the third follows by subtraction. If it asks for the determinant and lists the eigenvalues, multiply.

Properties You Must Recall Cold

Further standard results:

• A real symmetric matrix has all-real eigenvalues, and eigenvectors of distinct eigenvalues are orthogonal. This is why principal stresses are always real.
• A skew-symmetric real matrix has eigenvalues that are zero or purely imaginary.
• An orthogonal matrix has eigenvalues of unit modulus.
• $A$ is singular if and only if at least one eigenvalue is zero.
• Idempotent matrices ($A^2 = A$) have eigenvalues only $0$ or $1$.

Worked Example: Eigenvalues of a 2x2 Matrix

Take

$A = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}$

Trace $= 7$ and $\det A = 4(3) - 1(2) = 10$. The characteristic equation is therefore $\lambda^2 - 7\lambda + 10 = 0$, giving $\lambda = 5$ and $\lambda = 2$. Check: $5 + 2 = 7$ and $5 \times 2 = 10$.

For $\lambda = 5$, solve $(A - 5I)X = 0$:

$\begin{bmatrix} -1 & 1 \\ 2 & -2 \end{bmatrix}\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$

Both rows give $x_1 = x_2$, so the eigenvector is any non-zero multiple of $[1 ;; 1]^T$. For $\lambda = 2$ the same procedure gives $2x_1 + x_2 = 0$, i.e. $[1 ;; -2]^T$.

Notice that the eigenvector is only ever determined up to a scale factor — options that differ only by a constant multiple are the same answer, and CIL exploits this by listing scaled versions as distractors.

The Cayley-Hamilton Theorem

Every square matrix satisfies its own characteristic equation. For the matrix above, whose characteristic equation is $\lambda^2 - 7\lambda + 10 = 0$, the theorem asserts

$A^2 - 7A + 10I = 0$

This has two high-value uses.

Finding the inverse. Multiply through by $A^{-1}$:

$A - 7I + 10A^{-1} = 0 \quad \Longrightarrow \quad A^{-1} = \frac{1}{10}\left(7I - A\right)$

Substituting gives

$A^{-1} = \frac{1}{10}\begin{bmatrix} 3 & -1 \\ -2 & 4 \end{bmatrix}$

which you can verify equals the standard adjoint-over-determinant formula. For a $3\times3$ matrix this route is usually faster than computing nine cofactors.

Reducing high powers. From $A^2 = 7A - 10I$, any power collapses by repeated substitution: $A^3 = 7A^2 - 10A = 7(7A - 10I) - 10A = 39A - 70I$. A question asking for $A^4$ or $A^5$ is therefore arithmetic, not matrix multiplication.

Diagonalisation in One Line

If $A$ has $n$ linearly independent eigenvectors forming the columns of $P$, then $P^{-1}AP = D$, a diagonal matrix of eigenvalues. Consequently $A^k = PD^kP^{-1}$, and $D^k$ is just each eigenvalue raised to the power $k$. A matrix with $n$ distinct eigenvalues is always diagonalisable; repeated eigenvalues may or may not be.

Worked Example: A 3x3 Matrix End to End

CIL 3x3 items are usually built from small integer matrices carrying a repeated root, because that is exactly where candidates go wrong. Take

$A = \begin{bmatrix} 2 & 1 & 1 \\ 1 & 2 & 1 \\ 1 & 1 & 2 \end{bmatrix}$

Compute the three invariants rather than expanding the determinant symbolically. $S_1 = 2 + 2 + 2 = 6$. Each of the three principal $2\times2$ minors equals $2(2) - 1(1) = 3$, so $S_2 = 9$. Expanding once, $S_3 = \det A = 2(3) - 1(1) + 1(-1) = 4$. The characteristic equation is therefore

$\lambda^3 - 6\lambda^2 + 9\lambda - 4 = 0$

Test the small divisors of the constant term first. $\lambda = 1$ gives $1 - 6 + 9 - 4 = 0$, so $(\lambda - 1)$ divides the cubic and leaves $\lambda^2 - 5\lambda + 4 = (\lambda - 1)(\lambda - 4)$. The eigenvalues are $\lambda = 4, 1, 1$. Both shortcuts check out: $4 + 1 + 1 = 6 = S_1$, and $4 \times 1 \times 1 = 4 = \det A$.

For $\lambda = 4$, the system $(A - 4I)X = 0$ collapses to $x_1 = x_2 = x_3$, giving the eigenvector $[1 ;; 1 ;; 1]^T$. For $\lambda = 1$, $A - I$ is the all-ones matrix, so all three rows repeat the single condition $x_1 + x_2 + x_3 = 0$ — two free parameters, and therefore two independent eigenvectors such as $[1 ;; -1 ;; 0]^T$ and $[1 ;; 0 ;; -1]^T$. $A$ is real symmetric, and sure enough $[1 ;; 1 ;; 1]^T$ is orthogonal to both of them.

Algebraic Versus Geometric Multiplicity

A repeated eigenvalue does not by itself prevent diagonalisation. This is the single most misremembered rule in the topic, and CIL sets it directly.

• The algebraic multiplicity of an eigenvalue is the number of times it repeats as a root of the characteristic equation.
• The geometric multiplicity is the number of independent eigenvectors it yields, which equals $n - \text{rank}(A - \lambda I)$.
• Geometric multiplicity is always at least 1 and never exceeds the algebraic multiplicity. A matrix is diagonalisable exactly when the two agree for every eigenvalue.

In the worked example, $\lambda = 1$ has algebraic multiplicity 2, and $\text{rank}(A - I) = 1$ gives geometric multiplicity $3 - 1 = 2$. They agree, so $A$ is diagonalisable despite the repeat. Contrast that with

$B = \begin{bmatrix} 2 & 1 \\ 0 & 2 \end{bmatrix}$

whose only eigenvalue is 2 with algebraic multiplicity 2, while $\text{rank}(B - 2I) = 1$ leaves geometric multiplicity 1. $B$ is defective: there is no second independent eigenvector, so it cannot be diagonalised. A real symmetric matrix is never defective, which is why the symmetric case is always safe to assume diagonalisable.

Cayley-Hamilton on the 3x3

Applying the theorem to the same matrix gives $A^3 - 6A^2 + 9A - 4I = 0$, and multiplying through by $A^{-1}$ isolates the inverse:

$A^{-1} = \tfrac{1}{4}\left(A^2 - 6A + 9I\right)$

With $A^2$ equal to

$A^2 = \begin{bmatrix} 6 & 5 & 5 \\ 5 & 6 & 5 \\ 5 & 5 & 6 \end{bmatrix}$

this evaluates to

$A^{-1} = \frac{1}{4}\begin{bmatrix} 3 & -1 & -1 \\ -1 & 3 & -1 \\ -1 & -1 & 3 \end{bmatrix}$

Two independent checks confirm the result without multiplying anything out. The eigenvalues of $A^{-1}$ must be $1/4, 1, 1$, whose sum is $9/4$ — and the trace of the computed matrix is indeed $(3 + 3 + 3)/4 = 9/4$. Its determinant must be $1/\det A = 1/4$. Note that the method fails cleanly on a singular matrix: if $\det A = 0$ the constant term vanishes, so there is no $A^{-1}$ term to isolate, which is correct because no inverse exists.

Where This Returns Later in Paper-II

The same characteristic-equation machinery is reused across the Mechanical syllabus, so marks earned here are earned twice.

Because principal stresses are the eigenvalues of a symmetric tensor, they are guaranteed real and their planes mutually perpendicular. That is the theoretical justification for Mohr's circle which a purely graphical treatment never supplies.

Five Traps CIL Sets on This Topic

1. Offering the trace when the stem asks for the determinant, or the reverse. Both come from the same eigenvalue list in one line, so the wrong one is always sitting among the options.
2. Treating a repeated eigenvalue as automatic proof that a matrix cannot be diagonalised.
3. Forgetting that an eigenvector is fixed only up to scale, then rejecting the correct option because it is printed as a multiple of the vector you derived.
4. Extending "$A^T$ has the same eigenvalues as $A$" to the eigenvectors as well, which are generally different.
5. Reaching for Cayley-Hamilton to invert a singular matrix.