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Let $y$ be a non-zero vector of size $2022 \times 1.$ Which of the following statement(s) is/are $\text{TRUE}?$

  1. $yy^{T}$ is a symmetric matrix.
  2. $y^{T}y$ is an eigenvalue of $yy^{T}.$
  3. $yy^{T}$ has a rank of $2022.$
  4. $yy^{T}$ is invertible.

1 Answer

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  1. For any vector $y$, the transpose of the matrix product $yy^T$ is $(yy^T)^T = (y^T)^T y^T = yy^T$. Since the matrix equals its transpose, it is symmetric. (True)
  2. Consider the product $(yy^T)y = y(y^Ty)$. Since $y^Ty$ is a scalar, this fits the eigenvalue equation $Ax = \lambda x$ where $\lambda = y^Ty$. Thus, $y^Ty$ is an eigenvalue of $yy^T$. (True)
  3. The matrix $yy^T$ is formed by multiplying a column vector by a row vector. The rank of a product $AB$ is less than or equal to the minimum rank of $A$ or $B$. Since a non-zero vector has a rank of 1, the rank of $yy^T$ is exactly 1, not 2022. (False)
  4. A matrix is invertible only if it has full rank. Since the rank is 1 and the size is 2022, the matrix is singular (not invertible). (False)

 Options A and B 

 

Shortcut Trick: Any outer product matrix $uu^T$ always has a rank of 1 and its only non-zero eigenvalue is $u^Tu$. \

Important Rule: Rank of $(column \times row)$ is always 1 if the vectors are non-zero.

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