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In this step, we use the elementary matrix obtained by adding \(-3\) times the second row to the first row. The product \(AX_1\) is given by \[AX_1=\left ( \begin{array}{rrr} 2 & 2 & -2 \\ 1 & 3 & -1 \\ -1 & 1 & 1 \end{array} \right ) \left ( \begin{array}{r} 1 \\ 0 \\ 1 \end{array} \right ) = \left ( \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right )\]. Above relation enables us to calculate eigenvalues \lambda easily. \[\left ( \begin{array}{rrr} 5 & -10 & -5 \\ 2 & 14 & 2 \\ -4 & -8 & 6 \end{array} \right ) \left ( \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right ) = \left ( \begin{array}{r} -10 \\ 0 \\ 10 \end{array} \right ) =10\left ( \begin{array}{r} -1 \\ 0 \\ 1 \end{array} \right )\] This is what we wanted. Show that 2\\lambda is then an eigenvalue of 2A . Suppose the matrix \(\left(\lambda I - A\right)\) is invertible, so that \(\left(\lambda I - A\right)^{-1}\) exists. You should verify that this equation becomes \[\left(\lambda +2 \right) \left( \lambda +2 \right) \left( \lambda - 3 \right) =0\] Solving this equation results in eigenvalues of \(\lambda_1 = -2, \lambda_2 = -2\), and \(\lambda_3 = 3\). Compute \(AX\) for the vector \[X = \left ( \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right )\], This product is given by \[AX = \left ( \begin{array}{rrr} 0 & 5 & -10 \\ 0 & 22 & 16 \\ 0 & -9 & -2 \end{array} \right ) \left ( \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right ) = \left ( \begin{array}{r} 0 \\ 0 \\ 0 \end{array} \right ) =0\left ( \begin{array}{r} 1 \\ 0 \\ 0 \end{array} \right )\]. Definition \(\PageIndex{1}\): Eigenvalues and Eigenvectors, Let \(A\) be an \(n\times n\) matrix and let \(X \in \mathbb{C}^{n}\) be a nonzero vector for which. Let A =[2011]\begin{bmatrix}2 & 0\\-1 & 1\end{bmatrix}[2101], Example 3: Calculate the eigenvalue equation and eigenvalues for the following matrix , Let us consider, A =[100012200]\begin{bmatrix}1 & 0 & 0\\0 & -1 & 2\\2 & 0 & 0\end{bmatrix}102010020 All vectors are eigenvectors of I. This is unusual to say the least. \[\begin{aligned} X &=& IX \\ &=& \left( \left( \lambda I - A\right) ^{-1}\left(\lambda I - A \right) \right) X \\ &=&\left( \lambda I - A\right) ^{-1}\left( \left( \lambda I - A\right) X\right) \\ &=& \left( \lambda I - A\right) ^{-1}0 \\ &=& 0\end{aligned}\] This claims that \(X=0\). Eigenvalue is explained to be a scalar associated with a linear set of equations which when multiplied by a nonzero vector equals to the vector obtained by transformation operating on the vector. Notice that for each, \(AX=kX\) where \(k\) is some scalar. In the next example we will demonstrate that the eigenvalues of a triangular matrix are the entries on the main diagonal. The Mathematics Of It. We do this step again, as follows. : Find the eigenvalues for the following matrix? \[\left ( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 2 & 1 \end{array} \right ) \left ( \begin{array}{rrr} 33 & 105 & 105 \\ 10 & 28 & 30 \\ -20 & -60 & -62 \end{array} \right ) \left ( \begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -2 & 1 \end{array} \right ) =\left ( \begin{array}{rrr} 33 & -105 & 105 \\ 10 & -32 & 30 \\ 0 & 0 & -2 \end{array} \right )\] By Lemma [lem:similarmatrices], the resulting matrix has the same eigenvalues as \(A\) where here, the matrix \(E \left(2,2\right)\) plays the role of \(P\). If A is the identity matrix, every vector has Ax = x. In general, p i is a preimage of p i1 under A I. AI=[100012200]A \lambda I = \begin{bmatrix}1-\lambda & 0 & 0\\0 & -1-\lambda & 2\\2 & 0 & 0 \lambda \end{bmatrix}AI=102010020. Then \[\begin{array}{c} AX - \lambda X = 0 \\ \mbox{or} \\ \left( A-\lambda I\right) X = 0 \end{array}\] for some \(X \neq 0.\) Equivalently you could write \(\left( \lambda I-A\right)X = 0\), which is more commonly used. Let \[B = \left ( \begin{array}{rrr} 3 & 0 & 15 \\ 10 & -2 & 30 \\ 0 & 0 & -2 \end{array} \right )\] Then, we find the eigenvalues of \(B\) (and therefore of \(A\)) by solving the equation \(\det \left( \lambda I - B \right) = 0\). We often use the special symbol \(\lambda\) instead of \(k\) when referring to eigenvalues. In this case, the product \(AX\) resulted in a vector equal to \(0\) times the vector \(X\), \(AX=0X\). Thus the matrix you must row reduce is \[\left ( \begin{array}{rrr|r} 0 & 10 & 5 & 0 \\ -2 & -9 & -2 & 0 \\ 4 & 8 & -1 & 0 \end{array} \right )\] The is \[\left ( \begin{array}{rrr|r} 1 & 0 & - \vspace{0.05in}\frac{5}{4} & 0 \\ 0 & 1 & \vspace{0.05in}\frac{1}{2} & 0 \\ 0 & 0 & 0 & 0 \end{array} \right )\], and so the solution is any vector of the form \[\left ( \begin{array}{c} \vspace{0.05in}\frac{5}{4}s \\ -\vspace{0.05in}\frac{1}{2}s \\ s \end{array} \right ) =s\left ( \begin{array}{r} \vspace{0.05in}\frac{5}{4} \\ -\vspace{0.05in}\frac{1}{2} \\ 1 \end{array} \right )\] where \(s\in \mathbb{R}\). }\) The set of all eigenvalues for the matrix \(A\) is called the spectrum of \(A\text{.}\). These are the solutions to \(((-3)I-A)X = 0\). We need to show two things. The set of all eigenvalues of an \(n\times n\) matrix \(A\) is denoted by \(\sigma \left( A\right)\) and is referred to as the spectrum of \(A.\). 5. We wish to find all vectors \(X \neq 0\) such that \(AX = 2X\). 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