Linear algebra: rank, kernel and image explained

The first semester started a few days ago (at the Politecnico di Milano, lectures began on 14 September 2026), and in the middle of the first weeks of linear algebra and geometry there is already a concrete deadline on the horizon: the end-of-October mid-term test (prova in itinere), which at the Politecnico falls in the window 29 October – 2 November 2026, that is six weeks away. Behind it, the winter exam session: at universities with an early calendar like the Politecnico, the first exam date opens on 7 January 2027 and the session closes on 20 February. Every university has its own calendar, so check the course website — but the point does not change: the time to put things in order is now, not right before the test.

And what needs putting in order is rarely the calculations. Row reduction, determinants and matrix products are mechanical: you train them by repeating them, and they become automatic in a few weeks. What I see students fall on — in the oral exam, in a parametric system to discuss, in an exercise that is “different” from the ones seen in lectures — is the theoretical core: what a vector space is, what the rank really says, where the kernel and the image of a linear map live. These are things you can recite by heart without having understood them, and an examiner finds out in thirty seconds of questions.

The definitions that must be exact

In the oral exam, a definition given “roughly” is worth zero. Not because lecturers are pedantic for sport, but because in linear algebra every later theorem rests exactly on the words of the previous definition. If the definition is blurred, everything you build on top of it is just as blurred.

A vector space is a set $V$ on which addition and multiplication by a scalar are defined, with the properties that make these two operations “well behaved” (associativity, neutral element, distributivity…). In engineering courses you almost always work with ${\mathbb{R}}^n$, with the space of matrices or with that of polynomials: the abstract definition serves above all to recognise a subspace.

A subset $W\subseteq V$ is a subspace if and only if it satisfies three conditions, and they should be checked in this order:

1. it contains the zero vector ($\vec{0}\in W$);
2. it is closed under addition ($u,v\in W⟹u+v\in W$);
3. it is closed under multiplication by a scalar ($v\in W,\lambda \in \mathbb{R}⟹\lambda v\in W$).

Point 1 is also the fastest way to rule out that something is a subspace: if $\vec{0}$ does not satisfy it, you are done in five seconds, there is no need to check the other two points.

Linear independence: the vectors $v_1,\dots ,v_k$ are linearly independent if the only way to write

is with all the coefficients equal to zero ($c_1=c_2=\cdots =c_k=0$). Not “none is a multiple of another” (that holds only for two vectors), not “there are the right number of them”: it is exactly that condition on the linear combination.

A basis of $V$ is a set of vectors that spans (every vector of $V$ can be written as a linear combination of them) and is linearly independent. The dimension of $V$ is the number of vectors in any of its bases — and the “any” hides a non-obvious theorem: all the bases of the same space have the same number of elements. It is that theorem that makes dimension a well-defined number, not an opinion.

Rank, seen from three angles

The rank of a matrix $A$ is the most important concept of the course, and many students treat it superficially because it seems to be “just” the number of pivots after Gaussian elimination. That is true, but it is one of three equivalent descriptions, and knowing all three is what lets you answer any question however it is turned on you:

1. Number of pivots after row reduction — the operational definition, the one you actually use to compute it.
2. Maximum number of linearly independent rows (or columns) — the definition that explains why row reduction works: elementary operations do not change the space spanned by the rows, so they do not change how many rows are needed to span it.
3. Dimension of the image of $A$ seen as the linear map $T(x)=Ax$ — the definition that connects rank to the rest of the theory, the rank–nullity theorem included.

There is a fact that, the first time you see it, looks like a stroke of luck: the row rank is always equal to the column rank. It is not obvious a priori — rows and columns are different objects, they live in different spaces if the matrix is not square — and yet the maximum number of independent rows always coincides with the number of independent columns. It is a theorem, not a coincidence: being able to state it like that, and not as a little calculation rule, is one of the points where an examiner understands whether you have understood or are just running an algorithm.

Kernel and image: two sets that live in two different places

Given $T(x)=Ax$ with $A$ an $m\times n$ matrix, there are two fundamental subspaces, and the most common mistake — I talk about it below — is to swap them.

The kernel (or null space) is the set of vectors that $T$ sends to zero:

It lives in the domain, ${\mathbb{R}}^n$: it is literally the set of solutions of the homogeneous system associated with $A$. If $\dim (\ker T)=0$ the homogeneous system has only the trivial solution; if $\dim (\ker T)>0$ it has infinitely many, parametrised by that dimension.

The image is the set of vectors that $T$ reaches:

It lives in the codomain, ${\mathbb{R}}^m$, and coincides with the space spanned by the columns of $A$. Its dimension is, by definition, the rank of $A$.

The two subspaces are linked by the dimension theorem (or rank–nullity theorem):

where $\dim V$ is the dimension of the domain, that is $n$ (the number of columns of $A$). It is not just a result to quote: it is the most useful checking tool you have. Every time you compute a basis of the kernel and one of the image, add up their dimensions and check that they come back to $n$ — if they don’t, you have made a mistake in a calculation, even before you recheck it.

Rouché-Capelli reread through the rank

The Rouché-Capelli theorem is the translation, in terms of rank, of the question “does this system have a solution?”. Given the system $Ax=b$, with $A$ the coefficient matrix and $A|b$ the augmented matrix (with the column of constant terms added):

• the system is consistent if and only if $\mathrm{rk}(A)=\mathrm{rk}(A|b)$;
• if it is consistent, it has ${\infty }^{n-\mathrm{rk}(A)}$ solutions, where $n$ is the number of unknowns.

A consistent system with $\mathrm{rk}(A)=n$ has a unique solution, one with $\mathrm{rk}(A)<n$ has infinitely many, and the difference $n-\mathrm{rk}(A)$ is exactly $\dim (\ker A)$ — the same number as in the dimension theorem. They are not two unrelated formulas: it is the same idea seen from two sides.

The mistakes that cost marks

1. Confusing $\dim (\ker T)$ with $\dim (\mathrm{im}T)$. It happens when the dimension theorem has been learned as a formula and not as a relation between two different objects. Before applying it, ask yourself: which of the two have I already computed, and which one am I missing?

2. Swapping the domain with the codomain. The kernel lives in ${\mathbb{R}}^n$ (where the $x$ live), the image in ${\mathbb{R}}^m$ (where the $Ax$ live). If the matrix is not square they are spaces of different dimension, and swapping them is visible at a glance to whoever marks it.

3. Saying “the kernel is empty” instead of “it is the trivial subspace $\{\vec{0}\}$”. The kernel is never empty: it always contains zero, because $A\vec{0}=\vec{0}$ for any $A$. “Empty” and “reduced to zero alone” are different statements, and the first is false.

4. Declaring a matrix diagonalisable by looking only at the eigenvalues. It is the costliest trap in the programme — the next section is entirely about this.

5. Forgetting to check the case in which the parameter cancels exactly a pivot. In the parametric discussion the “special” value often does not jump out until you have row-reduced: it must be checked after the reduction, not estimated by eye beforehand.

Diagonalisability: the check that gets skipped

Here one of the most widespread and most dangerous mistakes is played out, because it seems harmless. If an $n\times n$ matrix has $n$ eigenvalues that are all distinct, then it is automatically diagonalisable — this is true, and convenient.

The problem is generalising the other way round: an eigenvalue with algebraic multiplicity greater than 1 guarantees nothing by itself. For every repeated eigenvalue you must explicitly check that

where the algebraic multiplicity is how many times the eigenvalue appears as a root of the characteristic polynomial, and the geometric one is the dimension of the associated eigenspace (that is $\dim (\ker (A-\lambda I))$ — a kernel again). It always holds that the geometric one is less than or equal to the algebraic one: if they are equal for every eigenvalue, the matrix is diagonalisable; if even for a single eigenvalue the geometric one is strictly smaller, it is not, full stop. Skipping this check — and concluding “diagonalisable” just because the characteristic polynomial factorises nicely — is the mistake I see cost the most marks of all. The third worked exercise below is built exactly on this case.

Three worked exercises

Exercise 1 — a parametric matrix and a discussion with Rouché-Capelli

Let

I compute the determinant by expanding along the first row:

I factorise (note that $k=1$ is a root): $k^3-3k+2=(k-1{)}^2(k+2)$. The determinant vanishes for $k=1$ (double) and $k=-2$: these are the values where the rank drops below 3.

I discuss the system $A(k)x=b$ with $b=(1,1,1{)}^T$.

For $k\notin \{1,-2\}$: $\det A(k)\ne 0$, so $\mathrm{rk}(A)=3=\mathrm{rk}(A|b)$. Consistent system, unique solution.

For $k=1$: $A(1)$ has all rows equal to $(1,1,1)$, so $\mathrm{rk}(A)=1$. The augmented matrix $(A|b)$ has all rows equal to $(1,1,1\mid 1)$, so $\mathrm{rk}(A|b)=1$ as well. The ranks coincide: consistent system, with ${\infty }^{3-1}={\infty }^2$ solutions, all those that satisfy $x+y+z=1$.

For $k=-2$: $A(-2)=\left(\begin{array}{ccc}-2& 1& 1\\ 1& -2& 1\\ 1& 1& -2\end{array}\right)$. The three rows add up to the zero vector, so they are dependent: $\mathrm{rk}(A)\le 2$, and the minor of the first two rows/columns equals $4-1=3\ne 0$, so $\mathrm{rk}(A)=2$ exactly. But the same dependence among the rows ($r_1+r_2+r_3=\vec{0}$) requires, for consistency, that the constant terms also add up to zero — here $1+1+1=3\ne 0$. So $\mathrm{rk}(A|b)=3\ne \mathrm{rk}(A)=2$: the system is inconsistent, no solution.

This is exactly the case “the parameter cancels a pivot” to keep an eye on: two special values of $k$, and two opposite outcomes — one consistent with infinitely many solutions, the other with no solution. If you train only on discussing the “generic” case and dismiss the special values as an appendix, that is where you lose full marks.

You can practise parametric discussions of this type, timed as in the exam, in the linear algebra practice test: the correct answer is free, the step-by-step guided solution unlocks with your first lesson.

Exercise 2 — kernel, image and checking the dimension theorem

Let

I notice at once that the second row is double the first, so $\mathrm{rk}(A)\le 2$; the first and third rows are not proportional, so $\mathrm{rk}(A)=2$.

Kernel. I solve $Ax=\vec{0}$. The second equation is a multiple of the first, so the system reduces to two independent equations:

Subtracting: $y+2z=0\Rightarrow y=-2z$. Substituting in the second: $x=-y-z=2z-z=z$. The kernel is therefore

Quick check: $A(1,-2,1{)}^T$ gives, row by row, $1-4+3=0$, $2-8+6=0$, $1-2+1=0$. It works.

Image. It is the space spanned by the columns of $A$: $c_1=(1,2,1)$, $c_2=(2,4,1)$, $c_3=(3,6,1)$. $c_2$ is not a multiple of $c_1$: you would need $c_2=2c_1$, which works on the first two components but on the third would give $2$ instead of $1$. The two columns are therefore independent and already form a basis of the image, consistent with $\mathrm{rk}(A)=2$:

(You can check that $c_3=-c_1+2c_2$, so it adds nothing new.)

Checking the dimension theorem: $\dim (\ker A)+\dim (\mathrm{im}A)=1+2=3=n$, the number of columns of $A$. It comes out exactly.

Exercise 3 — diagonalisability with a repeated eigenvalue

Let

Characteristic polynomial. I expand $\det (A-\lambda I)$ along the last row (it has only one non-zero entry):

Eigenvalues: ${\lambda }_1=2$ with algebraic multiplicity $2$, ${\lambda }_2=3$ with algebraic multiplicity $1$.

Eigenspace of ${\lambda }_2=3$. Algebraic multiplicity 1, so the geometric one is automatically 1: no check needed, equality is guaranteed.

Eigenspace of ${\lambda }_1=2$. I solve $(A-2I)v=\vec{0}$:

The equations are $y=0$ and $z=0$, with $x$ free. The eigenspace is

Geometric multiplicity of ${\lambda }_1$: 1. Algebraic multiplicity of ${\lambda }_1$: 2. They are different.

Conclusion: $A$ is not diagonalisable. It is not enough for the characteristic polynomial to factorise cleanly, and it is not even enough to have “only two distinct eigenvalues instead of three”: the matrix lacks an independent eigenvector exactly where two would be needed. If in an exercise you stop at computing the eigenvalues and declare it diagonalisable without having checked this eigenspace, you have answered the wrong question.

How we work in lessons for this exam

With linear algebra students I start by having them say the definition out loud, word by word, before touching an exercise: if a piece is missing there, any calculation done afterwards is built on sand. We work with the dimension theorem as a checking tool, not as a formula to quote from memory: before computing kernel and image I always ask “which dimension do you expect?”, and only afterwards do we check that the calculation confirms it.

We train the discussion of parametric systems separately from pure calculations: knowing how to row-reduce does not mean knowing how to discuss all the cases that the parameter can generate, including those that look like exceptions and are often instead the heart of the exam question. We work online with a graphics tablet and a shared screen, writing every step as on the blackboard in the oral exam.

For the general picture of how I work on maths there is the maths page; for support on first-year university courses, university tutoring. For a specific doubt before the mid-term test a single lesson is enough; if you have never had lessons with me there is the free 30-minute introductory call (updated rates on the pricing page). If in the same semester you also have numerical methods or control systems, linear algebra comes back there: the numerical solution of linear systems is in numerical methods exam, and the eigenvalues that decide the stability of a system are in control systems exam.

If you want to test yourself on the parametric discussion before the mid-term test, the linear algebra practice test has exercises in the same style as the ones seen here — correct answer free, guided solution with your first lesson. And if you are also putting analysis back in order, limits, linear systems and complex numbers are the chapters that intertwine most with this one. For any doubt, write to me: no pushy selling, just working out together where you stand.