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2026-01-01
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2026-02-28
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<p>In this section, we will be discussing the theorems on skew symmetric matrices. These theorems are useful in matrix decomposition, transformations, and applications. </p>
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<p>In this section, we will be discussing the theorems on skew symmetric matrices. These theorems are useful in matrix decomposition, transformations, and applications. </p>
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<p><strong>Theorem 1: For a square matrix A, A - AT is skew symmetric</strong></p>
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<p><strong>Theorem 1: For a square matrix A, A - AT is skew symmetric</strong></p>
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<p>For any square matrix A, we should prove that A - AT is skew symmetric.</p>
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<p>For any square matrix A, we should prove that A - AT is skew symmetric.</p>
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<p>We can take help from the properties of skew symmetric matrices, such as: </p>
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<p>We can take help from the properties of skew symmetric matrices, such as: </p>
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<ul><li>(A + B)T = AT + BT</li>
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<ul><li>(A + B)T = AT + BT</li>
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<li>(kA)T = kAT</li>
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<li>(kA)T = kAT</li>
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<li>(AT)T = A </li>
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<li>(AT)T = A </li>
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</ul><p>For A - AT to be skew symmetric, let’s prove that B = A - AT To prove B = A - AT, let’s take transpose on both sides. BT = (A - AT)T = AT - A BT = AT - (AT)T</p>
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</ul><p>For A - AT to be skew symmetric, let’s prove that B = A - AT To prove B = A - AT, let’s take transpose on both sides. BT = (A - AT)T = AT - A BT = AT - (AT)T</p>
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<p>We know that (AT)T = A Therefore, BT = AT - A BT = -(A - AT) BT = -B So, -B = -(A - AT)</p>
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<p>We know that (AT)T = A Therefore, BT = AT - A BT = -(A - AT) BT = -B So, -B = -(A - AT)</p>
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<p>We can now say that B = (A - AT) is skew-symmetric.</p>
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<p>We can now say that B = (A - AT) is skew-symmetric.</p>
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<p><strong>Theorem 2:</strong><strong>Decomposing a Square Matrix</strong></p>
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<p><strong>Theorem 2:</strong><strong>Decomposing a Square Matrix</strong></p>
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<p>Like all matrices, square matrices also encode complex relationships. It is important to decompose a square matrix into similar components to understand their properties and structure. A square matrix is expressed as the sum of a symmetric matrix, and a skew-symmetric matrix will be proved in this theorem. This decomposition is significant in applications like physics. </p>
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<p>Like all matrices, square matrices also encode complex relationships. It is important to decompose a square matrix into similar components to understand their properties and structure. A square matrix is expressed as the sum of a symmetric matrix, and a skew-symmetric matrix will be proved in this theorem. This decomposition is significant in applications like physics. </p>
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<p>In this theorem, we will be using the following properties:</p>
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<p>In this theorem, we will be using the following properties:</p>
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<ul><li>(AT)T = A</li>
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<ul><li>(AT)T = A</li>
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<li>(A + B)T = AT + BT</li>
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<li>(A + B)T = AT + BT</li>
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<li>(kA)T = kAT</li>
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<li>(kA)T = kAT</li>
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<li>If MT = M, then matrix M is symmetric</li>
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<li>If MT = M, then matrix M is symmetric</li>
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<li>If NT = -N, then matrix N is skew symmetric</li>
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<li>If NT = -N, then matrix N is skew symmetric</li>
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</ul><p>If A is a square matrix, it can be written as: A = \(\frac{1}{2}\) (A + AT) + 1/2 (A - AT) Let’s consider, P = \(\frac{1}{2}\) (A + AT) Q = \(\frac{1}{2}\) (A - AT)</p>
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</ul><p>If A is a square matrix, it can be written as: A = \(\frac{1}{2}\) (A + AT) + 1/2 (A - AT) Let’s consider, P = \(\frac{1}{2}\) (A + AT) Q = \(\frac{1}{2}\) (A - AT)</p>
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<p>Taking the transpose of both P and Q PT = (\(\frac{1}{2}\) (A + AT))T = ½(AT + (AT)T) = ½(AT + A) P = ½(AT + A) So, PT = P</p>
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<p>Taking the transpose of both P and Q PT = (\(\frac{1}{2}\) (A + AT))T = ½(AT + (AT)T) = ½(AT + A) P = ½(AT + A) So, PT = P</p>
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<p>QT = (\(\frac{1}{2}\) (A - AT))T = ½(AT - (AT)T) = ½ (AT - A) = ½ (AT - A) -Q = ½ (A - AT) So, QT = - Q</p>
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<p>QT = (\(\frac{1}{2}\) (A - AT))T = ½(AT - (AT)T) = ½ (AT - A) = ½ (AT - A) -Q = ½ (A - AT) So, QT = - Q</p>
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<p>So, the square matrix A can be written as the sum of a symmetric matrix P and a skew symmetric matrix Q. </p>
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<p>So, the square matrix A can be written as the sum of a symmetric matrix P and a skew symmetric matrix Q. </p>
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<p><strong>Theorem 3: For a skew symmetric matrix A and any matrix B, the matrix BTAB is skew symmetric.</strong></p>
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<p><strong>Theorem 3: For a skew symmetric matrix A and any matrix B, the matrix BTAB is skew symmetric.</strong></p>
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<p>Let’s take the help of the following properties:</p>
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<p>Let’s take the help of the following properties:</p>
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<ul><li>(AB)T = BTAT</li>
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<ul><li>(AB)T = BTAT</li>
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<li>(AT)T = A</li>
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<li>(AT)T = A</li>
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<li> AT = -A</li>
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<li> AT = -A</li>
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</ul><p>In this theorem, we shall prove (BTAB)T = -BTAB Start with: (BT AB)T = (BT (AB))T</p>
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</ul><p>In this theorem, we shall prove (BTAB)T = -BTAB Start with: (BT AB)T = (BT (AB))T</p>
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<p>Now use the transpose of a product rule: = (AB)T (BT)T</p>
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<p>Now use the transpose of a product rule: = (AB)T (BT)T</p>
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<p>Since (AB)T = BT AT we get: (BT AB)T = BTATB </p>
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<p>Since (AB)T = BT AT we get: (BT AB)T = BTATB </p>
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<p>Since AT = -A, we can make the substitution. </p>
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<p>Since AT = -A, we can make the substitution. </p>
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<p>Therefore, (BT AB)T = BT(-A)B (BT AB)T = -BT AB </p>
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<p>Therefore, (BT AB)T = BT(-A)B (BT AB)T = -BT AB </p>
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<p>So, if A is a skew symmetric matrix, then BT AB is a skew symmetric matrix. </p>
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<p>So, if A is a skew symmetric matrix, then BT AB is a skew symmetric matrix. </p>
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