Linear Algebra Domain Guide

Think of a matrix as a machine that takes a vector and returns another. Rank measures how much of the output space the machine can reach; nullity measures how much of the input space it destroys. Rank–nullity says those two numbers must add to the number of inputs — nothing disappears without being accounted for. This is why 'the columns are independent', 'the kernel is trivial', and 'the determinant is nonzero' are three statements of one fact for a square matrix.

Eigenvectors are the directions in which the machine is boring: it does not rotate or shear them, it only scales them. If you can find enough such directions to build a basis, the machine is, in the right coordinates, nothing but a list of scale factors — that is diagonalisation, and it is why $A^{100}$ becomes a computation on numbers instead of matrices.

The trace and the determinant are the cheap shadows of the spectrum: sum and product of the eigenvalues. On the $2\times 2$ and $3\times 3$ matrices this exam uses, they are usually enough to identify the eigenvalues by inspection, and they are always enough to catch an arithmetic slip.

Begin with the space, not the matrix. A subspace of $\mathbb{R}^n$ must contain $\vec{0}$ and be closed under addition and scaling — the fastest way to reject a candidate subspace on the exam is to check whether the zero vector satisfies the defining condition. Dimension counting then does most of the work: $\dim(U + W) = \dim U + \dim W - \dim(U \cap W)$, and for a linear map $T: V \to W$, $\dim V = \operatorname{rank}(T) + \operatorname{nullity}(T)$. Note the side that matters: nullity is counted against the dimension of the domain, so for an $m \times n$ matrix the sum is $n$, the number of columns, regardless of $m$.

The invertibility equivalences are worth memorising as one block, because a question phrased in any one of them can be answered from any other. For a square $n \times n$ matrix $A$, these are all equivalent: $A$ is invertible; $\det A \ne 0$; $\operatorname{rank} A = n$; the columns are linearly independent; the columns span $\mathbb{R}^n$; $A\vec{x} = \vec{0}$ has only the trivial solution; $A\vec{x} = \vec{b}$ has a unique solution for every $\vec{b}$; $0$ is not an eigenvalue of $A$; $A$ row-reduces to the identity.

Determinants are computed, but on this exam they are more often reasoned about. The properties do the work: $\det(AB) = \det A \det B$, $\det(A^T) = \det A$, $\det(A^{-1}) = 1/\det A$, a row swap flips the sign, adding a multiple of one row to another changes nothing, scaling one row by $c$ multiplies the determinant by $c$ — hence $\det(cA) = c^n \det A$ for an $n \times n$ matrix, which is the property most often mis-stated. A triangular matrix hands you its determinant as the product of the diagonal, which is why row-reducing to triangular form is the reliable computational route.

Eigenvalues come from $\det(A - \lambda I) = 0$. Two identities make them cheap: $\sum \lambda_i = \operatorname{tr}(A)$ and $\prod \lambda_i = \det(A)$. For a $2 \times 2$ matrix that is a complete system — the characteristic polynomial is $\lambda^2 - (\operatorname{tr} A)\lambda + \det A$. Once the eigenvalues of $A$ are known, the spectra of related matrices follow without new work: $A^k$ has eigenvalues $\lambda^k$, $A^{-1}$ has $1/\lambda$, $A + cI$ has $\lambda + c$, and $A^T$ has the same eigenvalues as $A$ (though generally different eigenvectors). A polynomial $p$ applied to $A$ has eigenvalues $p(\lambda)$.

Diagonalisability is the question students most often answer with the wrong rule. Distinct eigenvalues are sufficient — eigenvectors for distinct eigenvalues are automatically independent — but they are not necessary: the identity matrix has one eigenvalue repeated $n$ times and is already diagonal. The correct criterion is that for every eigenvalue the geometric multiplicity $\dim\ker(A - \lambda I)$ equals the algebraic multiplicity. When it fails, the matrix is defective and its Jordan form has a block larger than $1 \times 1$. Real symmetric matrices never fail: they have real eigenvalues, and eigenvectors for distinct eigenvalues are orthogonal, so they are orthogonally diagonalisable.

Linear operators are the same content stated basis-free, and the exam uses that framing for questions about polynomial spaces and function spaces. If $T$ is differentiation on the space of polynomials of degree at most $3$, then in the basis $\{1, x, x^2, x^3\}$ it is a strictly upper-triangular matrix: its only eigenvalue is $0$, its kernel is the constants (nullity $1$), and its rank is $3$ — the entire structure of the operator is read off from a matrix you can write down in a few seconds. Two operators that are similar are the same operator seen from two bases, so any quantity that depends only on the operator — rank, trace, determinant, characteristic polynomial, eigenvalues — must agree.