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https://texercises.com/exercise/diagonalizable-matrices-and-eigenvectors/
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Exercise:
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.

Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
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Exercise:
T is diagonalizable iff exists a basis mathcalBv_...v_n consisting of eigenvectors of T i.e. forall i Tv_ilambda_i v_i for some lambda_iin K. Moreover if for some basis mathcalBv_...v_n the matrix T_mathcalB^mathcalB has the shape pmatrix lambda_ & hdots & vdots & lambda_i & vdots & hdots & lambda_n pmatrix. then forall i lambda_i is an eigenvalue of T and v_i is an eigenvector for lambda_i.

Solution:
Proof. Recall that T_mathcalB^mathcalB pmatrix vdots & & vdots Tv__mathcalB & hdots & Tv_n_mathcalB vdots & & vdots pmatrix. and Tv_j_mathcalB pmatrix alpha_j hdots alpha_nj pmatrix where Tv_jalpha_jv_+...+alpha_njv_n. T_mathcalB^mathcalB has the shape above iff alpha_ijquad forall ineq j and alpha_kklambda_kquad k iff Tv_k lambda_k v_k.
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eigenvalue, eigenvector, eth, fs23, lineare algebra, matrices, proof
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(3, default)
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ENG (English)
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Proof
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