IIT JAM Exam > IIT JAM Questions > Let Pbe an nxnnon-null real skew-symmetric ma...
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Let P be an n x n non-null real skew-symmetric matrix, where n is even. Which of the following statements is (are) always TRUE?
- a)Px = 0 has infinitely many solutions, where 0 ∈ ℝn
- b)Px = λx has a unique solution for every non-zero λ ∈ ℝ
- c)If Q = (In + P)(In − P)−1, then QT Q = In
- d)The sum of all the eigenvalues of P is zero
Correct answer is option 'B,C,D'. Can you explain this answer?
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Most Upvoted Answer
Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Whi...
Since $P$ is skew-symmetric, we have $P^T = -P.$ Taking the transpose of both sides of $Px = \lambda x,$ we get
\[x^T P^T = \overline{\lambda} x^T,\]so
\[-x^T P = \overline{\lambda} x^T.\]Multiplying both sides by $x,$ we get
\[-x^T Px = \overline{\lambda} x^T x,\]so $\lambda$ is purely imaginary. Furthermore, $x^T Px = 0,$ or $x^T P^T x = -(x^T P x) = 0.$ Then for any vector $y,$
\begin{align*}
y^T P^T x &= -(y^T P x) \\
&= -(x^T P^T y) \\
&= x^T P^T y \\
&= -x^T P y.
\end{align*}Then $y^T P^T x = -y^T P x$ for all vectors $y,$ which forces $Px = 0.$ Thus, the only eigenvalues of $P$ are 0 and purely imaginary numbers.
(a) Suppose $Px = 0.$ Then $x^T P x = 0,$ so every eigenvector corresponding to a nonzero eigenvalue of $P$ must have a 0 entry. In particular, the eigenvectors corresponding to the purely imaginary eigenvalues must have a 0 entry.
Let $v$ be an eigenvector corresponding to a nonzero eigenvalue of $P.$ Then $Pv$ is an eigenvector corresponding to the same eigenvalue, so
\[v^T P^T v = v^T (-P) v = -v^T P v,\]which means $v^T P v = 0.$ Then $v$ has a 0 entry. Hence, the only eigenvector of $P$ that corresponds to 0 is the zero vector.
Therefore, $\det P = 0,$ so $\det (P - \lambda I) = \lambda^n,$ which means the eigenvalues of $P$ must be 0 or purely imaginary. Since the determinant of $P$ is 0, the product of the eigenvalues is 0, which means there must be at least two 0 eigenvalues. Since the matrix is non-null, there must be at least two nonzero eigenvalues, which means there must be at least two purely imaginary eigenvalues.
Let $v$ be an eigenvector corresponding to a purely imaginary eigenvalue $\lambda = bi,$ where $b \neq 0.$ Then
\[v^T P^T v = (-v^T P v) = -\lambda \|v\|^2 = -b^2 \|v\|^2,\]so $\|v\| = 0,$ which means $v = 0.$ Therefore, the only eigenvalue of $P$ is 0, so $P$ is similar to a direct sum of matrices of the form
\[\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\]The matrix $P$ is diagonalizable, so
\[x^T P^T = \overline{\lambda} x^T,\]so
\[-x^T P = \overline{\lambda} x^T.\]Multiplying both sides by $x,$ we get
\[-x^T Px = \overline{\lambda} x^T x,\]so $\lambda$ is purely imaginary. Furthermore, $x^T Px = 0,$ or $x^T P^T x = -(x^T P x) = 0.$ Then for any vector $y,$
\begin{align*}
y^T P^T x &= -(y^T P x) \\
&= -(x^T P^T y) \\
&= x^T P^T y \\
&= -x^T P y.
\end{align*}Then $y^T P^T x = -y^T P x$ for all vectors $y,$ which forces $Px = 0.$ Thus, the only eigenvalues of $P$ are 0 and purely imaginary numbers.
(a) Suppose $Px = 0.$ Then $x^T P x = 0,$ so every eigenvector corresponding to a nonzero eigenvalue of $P$ must have a 0 entry. In particular, the eigenvectors corresponding to the purely imaginary eigenvalues must have a 0 entry.
Let $v$ be an eigenvector corresponding to a nonzero eigenvalue of $P.$ Then $Pv$ is an eigenvector corresponding to the same eigenvalue, so
\[v^T P^T v = v^T (-P) v = -v^T P v,\]which means $v^T P v = 0.$ Then $v$ has a 0 entry. Hence, the only eigenvector of $P$ that corresponds to 0 is the zero vector.
Therefore, $\det P = 0,$ so $\det (P - \lambda I) = \lambda^n,$ which means the eigenvalues of $P$ must be 0 or purely imaginary. Since the determinant of $P$ is 0, the product of the eigenvalues is 0, which means there must be at least two 0 eigenvalues. Since the matrix is non-null, there must be at least two nonzero eigenvalues, which means there must be at least two purely imaginary eigenvalues.
Let $v$ be an eigenvector corresponding to a purely imaginary eigenvalue $\lambda = bi,$ where $b \neq 0.$ Then
\[v^T P^T v = (-v^T P v) = -\lambda \|v\|^2 = -b^2 \|v\|^2,\]so $\|v\| = 0,$ which means $v = 0.$ Therefore, the only eigenvalue of $P$ is 0, so $P$ is similar to a direct sum of matrices of the form
\[\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.\]The matrix $P$ is diagonalizable, so
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Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer?
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Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer?.
Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer? for IIT JAM 2024 is part of IIT JAM preparation. The Question and answers have been prepared according to the IIT JAM exam syllabus. Information about Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer? covers all topics & solutions for IIT JAM 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let Pbe an nxnnon-null real skew-symmetric matrix, where nis even. Which of the following statements is (are) always TRUE?a) Px= 0has infinitely many solutions, where 0 ∈ nb) Px= λxhas a unique solution for every non-zero λ∈ c) If Q= (In+ P)(In− P)−1, then QTQ= Ind) The sum of all the eigenvalues of Pis zeroCorrect answer is option 'B,C,D'. Can you explain this answer?.
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