QS-019 · Linear Algebra · Easy

GATE DA 2025 | Question: 40

Let $x_1,x_2,x_3,x_4,x_5$ be a system of orthonormal vectors in $\mathbb{R}^{10}$. Consider the matrix $$ A=x_1x_1^\top+x_2x_2^\top+x_3x_3^\top+x_4x_4^\top+x_5x_5^\top. $$ Which of the following statements is/are correct? 1. Singular values of $A$ are also its eigenvalues 2. Singular values of $A$ are either $0$ or $1$ 3. Determinant of $A$ is $1$ 4. $A$ is invertible

Solution Let $X=[x_1\ x_2\ x_3\ x_4\ x_5]$. Since the vectors are orthonormal, $X^\top X=I_5$ and $A=XX^\top$. Therefore, $$ A^2=XX^\top XX^\top=X(X^\top X)X^\top=XX^\top=A. $$ So $A$ is symmetric and idempotent. For each $i=1,\ldots,5$, $Ax_i=x_i$, so 1 is an eigenvalue with multiplicity 5. The orthogonal complement of $\operatorname{span}{x_1,\ldots,x_5}$ has dimension 5, and $Av=0$ for every vector in this complement. Hence, the eigenvalues are 1 (five times) and 0 (five times). Since $A^\top A=A^2=A$, the singular values are also 1 (five times) and 0 (five times). Thus, **Statement 1 and Statement 2 are true**. Since $A$ has zero eigenvalues, $\det(A)=0$, so $A$ is not invertible. Therefore, **Statements 3 and 4 are false**. **Correct options: 1 and 2.**

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