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The Jordan canonical form of a real symmetric matrix with characteristic polynomial \((x - 1)^{4}\) is :

  • A.\[ \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
  • B.\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
  • C.\[ \begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]
  • D.\[ \begin{bmatrix} 1 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \]

Answer: A

The Jordan form is the 4 × 4 identity matrix.

A real symmetric matrix is orthogonally diagonalizable (the spectral theorem), so its Jordan canonical form is diagonal, with its eigenvalues on the diagonal. The characteristic polynomial \((x - 1)^{4}\) says the only eigenvalue is 1, with algebraic multiplicity 4. A diagonal matrix with four 1s is I. In fact the symmetric matrix itself must be I, since \(Q I Q^{T} = I\).

The other three matrices have 1s above the diagonal, that is, Jordan blocks of size 2, 4 and so on, and such matrices are not diagonalizable. The last one is not even in Jordan form, since its off-diagonal 1 is not on the superdiagonal.

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