The Transpose of a Nonsingular Matrix is Nonsingular

Linear algebra problems and solutions

Problem 558

Let $A$ be an $n\times n$ nonsingular matrix.

Prove that the transpose matrix $A^{\trans}$ is also nonsingular.

 
LoadingAdd to solve later

Sponsored Links


Definition (Nonsingular Matrix).

By definition, $A^{\trans}$ is a nonsingular matrix if the only solution to
\[A^{\trans}\mathbf{x}=\mathbf{0}\] is the zero vector $\mathbf{x}=\mathbf{0}$ in $\R^n$.

Proof.

Suppose that
\[A^{\trans}\mathbf{v}=\mathbf{0} \tag{*}\] for an $n$-dimensional vector $\mathbf{v}$.
We prove that $\mathbf{v}=\mathbf{0}$.


Since $A$ is nonsingular, there exists a vector $\mathbf{u}\in \R^n$ such that
\[A\mathbf{u}=\mathbf{v}.\] (For those who know the inverse matrix, the vector $\mathbf{u}$ is given by $\mathbf{u}=A^{-1}\mathbf{v}$. The fact can be proved without using the inverse matrix, though.)

Hence we obtain from (*)
\[A^{\trans}A\mathbf{u}=\mathbf{0}.\] It follows that
\[\mathbf{u}^{\trans}A^{\trans}A\mathbf{u}=\mathbf{u}^{\trans}\mathbf{0}=0.\] By the property of the transpose, we have
\[\mathbf{u}^{\trans}A^{\trans}=(A\mathbf{u})^{\trans}.\] Thus we have
\[0=(A\mathbf{u})^{\trans}A\mathbf{u}=\|A\mathbf{u}\|^2.\] This yields the length $\|A\mathbf{u}\|=0$, and hence $A\mathbf{u}=\mathbf{0}$.
Since $\mathbf{v}=A\mathbf{u}$, we conclude that $\mathbf{v}=\mathbf{0}$.


Therefore if $A^{\trans}\mathbf{v}=\mathbf{0}$, then $\mathbf{v}=\mathbf{0}$.
This shows that the transpose $A^{\trans}$ is nonsingular.


LoadingAdd to solve later

Sponsored Links

More from my site

  • Construction of a Symmetric Matrix whose Inverse Matrix is ItselfConstruction of a Symmetric Matrix whose Inverse Matrix is Itself Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the $n\times n$ matrix $A$ by \[A=I-a\mathbf{v}\mathbf{v}^{\trans},\] where […]
  • Find the Inverse Matrix of a Matrix With FractionsFind the Inverse Matrix of a Matrix With Fractions Find the inverse matrix of the matrix \[A=\begin{bmatrix} \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\[6 pt] \frac{6}{7} &\frac{2}{7} &-\frac{3}{7} \\[6pt] -\frac{3}{7} & \frac{6}{7} & -\frac{2}{7} \end{bmatrix}.\]   Hint. You may use the augmented matrix […]
  • Subspaces of Symmetric, Skew-Symmetric MatricesSubspaces of Symmetric, Skew-Symmetric Matrices Let $V$ be the vector space over $\R$ consisting of all $n\times n$ real matrices for some fixed integer $n$. Prove or disprove that the following subsets of $V$ are subspaces of $V$. (a) The set $S$ consisting of all $n\times n$ symmetric matrices. (b) The set $T$ consisting of […]
  • Diagonalizable by an Orthogonal Matrix Implies a Symmetric MatrixDiagonalizable by an Orthogonal Matrix Implies a Symmetric Matrix Let $A$ be an $n\times n$ matrix with real number entries. Show that if $A$ is diagonalizable by an orthogonal matrix, then $A$ is a symmetric matrix.   Proof. Suppose that the matrix $A$ is diagonalizable by an orthogonal matrix $Q$. The orthogonality of the […]
  • Compute and Simplify the Matrix Expression Including Transpose and Inverse MatricesCompute and Simplify the Matrix Expression Including Transpose and Inverse Matrices Let $A, B, C$ be the following $3\times 3$ matrices. \[A=\begin{bmatrix} 1 & 2 & 3 \\ 4 &5 &6 \\ 7 & 8 & 9 \end{bmatrix}, B=\begin{bmatrix} 1 & 0 & 1 \\ 0 &3 &0 \\ 1 & 0 & 5 \end{bmatrix}, C=\begin{bmatrix} -1 & 0\ & 1 \\ 0 &5 &6 \\ 3 & 0 & […]
  • Find the Distance Between Two Vectors if the Lengths and the Dot Product are GivenFind the Distance Between Two Vectors if the Lengths and the Dot Product are Given Let $\mathbf{a}$ and $\mathbf{b}$ be vectors in $\R^n$ such that their length are \[\|\mathbf{a}\|=\|\mathbf{b}\|=1\] and the inner product \[\mathbf{a}\cdot \mathbf{b}=\mathbf{a}^{\trans}\mathbf{b}=-\frac{1}{2}.\] Then determine the length $\|\mathbf{a}-\mathbf{b}\|$. (Note […]
  • Eigenvalues of a Matrix and its Transpose are the SameEigenvalues of a Matrix and its Transpose are the Same Let $A$ be a square matrix. Prove that the eigenvalues of the transpose $A^{\trans}$ are the same as the eigenvalues of $A$.   Proof. Recall that the eigenvalues of a matrix are roots of its characteristic polynomial. Hence if the matrices $A$ and $A^{\trans}$ […]
  • Rank and Nullity of a Matrix, Nullity of TransposeRank and Nullity of a Matrix, Nullity of Transpose Let $A$ be an $m\times n$ matrix. The nullspace of $A$ is denoted by $\calN(A)$. The dimension of the nullspace of $A$ is called the nullity of $A$. Prove the followings. (a) $\calN(A)=\calN(A^{\trans}A)$. (b) $\rk(A)=\rk(A^{\trans}A)$.   Hint. For part (b), […]

You may also like...

Leave a Reply

Your email address will not be published. Required fields are marked *

This site uses Akismet to reduce spam. Learn how your comment data is processed.

More in Linear Algebra
Inverse Matrices Problems and Solutions
Construction of a Symmetric Matrix whose Inverse Matrix is Itself

Let $\mathbf{v}$ be a nonzero vector in $\R^n$. Then the dot product $\mathbf{v}\cdot \mathbf{v}=\mathbf{v}^{\trans}\mathbf{v}\neq 0$. Set $a:=\frac{2}{\mathbf{v}^{\trans}\mathbf{v}}$ and define the...

Close