Proof that the kernel of a linear map is a subspace

Primary Tool

Proof Builder

This proof is easiest to follow when the three subspace checks stay separate. The layout keeps the zero-vector step, the addition step, and the scalar-multiplication step visible at the same time.

The subspace test in action

To prove $ker (T)$ is a subspace of $V$ , you do not need a special theorem beyond the subspace test. You need to show $0_{V} \in ker (T)$ and that the kernel is closed under addition and scalar multiplication.

The proof is short because linearity already packages the needed algebra: $T (u + v) = T (u) + T (v)$ and $T (c u) = c T (u)$ .

Where students usually lose the thread

The definition of kernel only says which vectors land at $0_{W}$ . It does not by itself tell you closure. Closure comes from applying linearity to vectors you already know lie in the kernel.

That is why the proof alternates between a structural step, like 'let $u, v \in ker (T)$ ', and an algebraic step, like 'then $T (u + v) = 0_{W}$ '.

Common Pitfall

Checking addition alone is not enough. To prove the kernel is a subspace, you still need to show that $0_{V}$ is in the kernel and that scalar multiples stay in the kernel too.

Try a Variation

Try the same template for the image of a linear map. Which part of the subspace test becomes slightly different, and why?

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ExplanationLinear AlgebraMath Workspace (CAS)

What eigenvectors and eigenvalues mean geometrically

An eigenvector is a direction that survives a linear transformation without rotating away from its own line. The eigenvalue tells you the scale and possible sign flip along that direction.

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Worked ExampleLinear AlgebraMath Workspace (CAS)

Diagonalize a $2 \times 2$ matrix and compute $A^{n}$

Diagonalization is one of the first places students see why eigenvalues matter computationally. Once $A = P D P^{- 1}$ , powers of $A$ become powers of a diagonal matrix.

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