Video summary

Eigen values and Eigen vectors - RANK OF MATRIX

Main summary

Key takeaways

Educational

Main ideas / concepts conveyed

  • Eigenvalue–eigenvector problem setup

    • For a square matrix (A), eigenvectors are special vectors whose direction does not change under the linear transformation by (A).
    • The vector may change length/scale by a factor called the eigenvalue.
    • If (X) is an eigenvector and (\lambda) is the eigenvalue, then: [ AX=\lambda X \quad\Longleftrightarrow\quad (A-\lambda I)X=0 ]

  • Why determinant matters

    • ((A-\lambda I)X=0) is a homogeneous system.
    • It always has the trivial solution (X=0), but the goal is non-trivial (infinite) solutions, which occur when the system is singular.

    • The condition for non-trivial solutions:

      • [ \det(A-\lambda I)=0 ]

      • This yields a polynomial equation in (\lambda): the characteristic equation.

  • Characteristic equation and degree

    • If (A) is:
      • (2\times 2) → characteristic equation is quadratic
      • (3\times 3) → characteristic equation is cubic
      • (and so on)

  • How to proceed once eigenvalues are known

    • For each eigenvalue (\lambda), solve: [ (A-\lambda I)X=0 ]

    • This gives the corresponding eigenvector(s).

Method / instructions presented (detailed step list)

A) Find eigenvalues (via characteristic equation)

  1. Start with the matrix (A).
  2. Form (A-\lambda I) by subtracting (\lambda) from the diagonal entries of (A).

  3. Compute: [ \det(A-\lambda I)=0 ]

  4. Expand/simplify to get the characteristic equation (a polynomial in (\lambda)).

  5. Solve the polynomial equation to obtain the eigenvalues.

B) Find eigenvectors for each eigenvalue

For each computed eigenvalue (\lambda):

  • Form the homogeneous system: [ (A-\lambda I)X=0 ]

  • Write the matrix equation as a system in unknowns (X_1, X_2, X_3) (for a (3\times 3) matrix).

  • Use the “rule of cross multiplication” shortcut:
    • Choose any two rows (the video repeatedly uses the 2nd and 3rd rows).
    • Express the ratios (X_1, X_2, X_3) using determinants of (2\times 2) minors.
    • The result gives eigenvectors up to a multiplicative constant (K).
  • Simplify by cancelling common factors.
  • The eigenvector is the resulting vector (direction matters; scaling does not).

Worked example in the video (main results)

Given matrix

[ A=\begin{bmatrix} 2 & 2 & 0\ 2 & 1 & 1\ -7 & 2 & -3 \end{bmatrix} ]

Step 1: Characteristic equation

  • Form (A-\lambda I).

  • Set determinant to zero: [ \det(A-\lambda I)=0 ]

  • The video arrives at: [ \lambda^3 - 13\lambda + 12 = 0 ] (Equivalently written in the video as ( \lambda^2 - 13\lambda + 12 = 0) after intermediate rewriting; the final eigenvalues are obtained using a calculator.)

  • Eigenvalues (as stated): [ \lambda=-\frac{4}{3},\; 1,\; \frac{1}{2} ] (The video transcript is somewhat garbled around numerical formatting, but these are the values it concludes with.)

Step 2: Eigenvectors (as computed in the video)

The video computes eigenvectors using the cross-multiplication shortcut:

  • For (\lambda=-4) (shown in the worked eigenvector section): [ X=\begin{bmatrix}1\-3\13\end{bmatrix} ]

  • For (\lambda=3): [ X=\begin{bmatrix}2\1\-2\end{bmatrix} ]

  • For (\lambda=1): [ X=\begin{bmatrix}-2\1\4\end{bmatrix} ]

Note: The eigenvector computations shown are explicitly carried out for (\lambda=-4, 3, 1), which conflicts with the earlier stated eigenvalues in the transcript. This likely comes from transcript/auto-subtitle errors.

Speakers / sources featured

  • Susan and John (channel names mentioned in the greeting)
  • The video narrator/instructor (unnamed; speaks throughout)
  • No other sources/guests are referenced

Original video