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If i given eigenvector: $$V_1=\begin{pmatrix} {1\over \sqrt{3}}\\{1\over \sqrt{3}}\\{1\over \sqrt{3}}\end{pmatrix} , V_2=\begin{pmatrix} {1\over \sqrt{6}}\\{-2\over \sqrt{6}}\\{1\over \sqrt{6}}\end{pmatrix} Space:\lambda=1 \\ V_3=\begin{pmatrix} {1\over \sqrt{2}}\\0\\{1\over \sqrt{2}}\end{pmatrix}Space: \lambda = 3$$

  • I know finding eigenvectors from matrix but how can i find Matix from those given vectors and $\lambda$s ?
  • Are those vectors orthonormal? (what are the properties of orthonormality ?)

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      $\begingroup$ You can't recover the matrix from the eigenpairs alone. $\endgroup$ – Git Gud Jan 4 '15 at 20:39
    • $\begingroup$ But i asked to find inverse of the matrix from it at the exam $\endgroup$ – Andy Jan 4 '15 at 20:40
    • $\begingroup$ I suggest you post the exact problem, maybe there's a misunderstanding somewhere. $\endgroup$ – Git Gud Jan 4 '15 at 20:41
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      $\begingroup$ Since it is an exam question, I suspect the third vector was $(\pm 1,0,\mp 1)/\sqrt{2}$. That would give 3 orthonormal vectors of $A$. The inverse matrix, $A^{-1}$, would then be given by $1/\lambda_1 V_1 V_1^T + 1/\lambda_2 V_2 V_2^T + 1/\lambda_3 V_3 V_3^T$. Note that I am using $V V^T$ to denote the outer product of two vectors. $\endgroup$ – LouisB Jan 4 '15 at 22:49
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      $\begingroup$ @Andy Here is a link to matrix inverse eigendecomposition. You'll have to scroll down a few sections. The formula $A = Q \Lambda Q^T$ is known as the spectral theorem or the principal axis theorem. $Q$ is a matrix whose columns are the eigenvectors of $A$. The expansion that I used is a technique used for constructing a matrix from projections or from projection operators. Gilbert Strang uses it in Linear Algebra and Its Applications, 4th edition, Section 5.5. $\endgroup$ – LouisB Jan 5 '15 at 0:30
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    you can see that the vectors $\{v_1, v_2, v_3 \}$ is orthonormal meaning each has unit length($v_1.v_1 = v_1^T v_1 = 1$) and any pair($v_1.v_2 = v_1^Tv_2 = 0$) is orthogonal. so they for a basis for $R^3$ if you compose $A = v_1v_1^T + v_2v_2^T + 3v_3v_3^T$ as the linear combination three rank one matrices, then you can verify that $Av_1 = v_1, Av_2 = v_2$ and $Av_3 = 3v_3$ for what this tells us is that the matrix has eigenvalues $1,1,3$ and the corresponding eigenvectors $v_1, v_2, v_3$

    in fact, if $B = kv_1v_1^T + lv_2v_2^T + mv_3v_3^T,$ then the matrix $B$ has eigenvalues $k,l,m$ we can use this fact and that $A^{-1}$ has eigenvalues $1, 1, 1/3$(reciprocals of the eigenvalues of $A$) you have $$ A^{-1} = v_1v_1^T + v_2v_2^T + {1 \over3}v_3v_3^T$$

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