Exercise 2171Linear algebra › Inner product spacesL1Added 3 Nov 2026

Orthogonal projection onto a column space

#orthogonal projection#column space#normal equations

Statement

Let $A$ be an $m\times n$ matrix with full column rank ( $\operatorname{rank}A=n$ ). Show that the matrix of orthogonal projection onto $\operatorname{Col}(A)$ is
$P=A(A^TA)^{-1}A^T.$

Hint

Show that $P$ is symmetric, idempotent, and fixes $\operatorname{Col}(A)$ , and that its kernel is $\operatorname{Col}(A)^\perp$ .

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Solution

Symmetry. $P^T=(A^TA)^{-1}$ -part: $P^T=A((A^TA)^{-1})^T A^T=A(A^TA)^{-1}A^T=P$ since $A^TA$ is symmetric.

Idempotence.
$P^2=A(A^TA)^{-1}A^T A(A^TA)^{-1}A^T=A(A^TA)^{-1}(A^TA)(A^TA)^{-1}A^T=P.$

Fixes $\operatorname{Col}(A)$ . For $x=Ay$ : $Px=A(A^TA)^{-1}A^TAy=Ay=x$ .

Kernel. If $Px=0$ , then $A(A^TA)^{-1}A^T x=0$ . Multiply by $(A^TA)^{-1}A^T$ ? Actually, $\langle Px,x\rangle=0\iff\|A^T x\|^2_{?}=0$ . Note $x^T P x=x^TA(A^TA)^{-1}A^Tx=\|(A^TA)^{-1/2}A^Tx\|^2\ge0$ . If $Px=0$ , then $A^Tx=0$ , so $x\perp\operatorname{Col}(A)$ .

Conclusion. $P$ is the orthogonal projection onto $\operatorname{Col}(A)$ .

Orthogonal projection onto a column space

Statement

Solution

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