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Linear Algebra Examples
[01-1√2]
Step 1
Set up the formula to find the characteristic equation p(λ).
p(λ)=determinant(A-λI2)
Step 2
The identity matrix or unit matrix of size 2 is the 2×2 square matrix with ones on the main diagonal and zeros elsewhere.
[1001]
Step 3
Step 3.1
Substitute [01-1√2] for A.
p(λ)=determinant([01-1√2]-λI2)
Step 3.2
Substitute [1001] for I2.
p(λ)=determinant([01-1√2]-λ[1001])
p(λ)=determinant([01-1√2]-λ[1001])
Step 4
Step 4.1
Simplify each term.
Step 4.1.1
Multiply -λ by each element of the matrix.
p(λ)=determinant([01-1√2]+[-λ⋅1-λ⋅0-λ⋅0-λ⋅1])
Step 4.1.2
Simplify each element in the matrix.
Step 4.1.2.1
Multiply -1 by 1.
p(λ)=determinant([01-1√2]+[-λ-λ⋅0-λ⋅0-λ⋅1])
Step 4.1.2.2
Multiply -λ⋅0.
Step 4.1.2.2.1
Multiply 0 by -1.
p(λ)=determinant([01-1√2]+[-λ0λ-λ⋅0-λ⋅1])
Step 4.1.2.2.2
Multiply 0 by λ.
p(λ)=determinant([01-1√2]+[-λ0-λ⋅0-λ⋅1])
p(λ)=determinant([01-1√2]+[-λ0-λ⋅0-λ⋅1])
Step 4.1.2.3
Multiply -λ⋅0.
Step 4.1.2.3.1
Multiply 0 by -1.
p(λ)=determinant([01-1√2]+[-λ00λ-λ⋅1])
Step 4.1.2.3.2
Multiply 0 by λ.
p(λ)=determinant([01-1√2]+[-λ00-λ⋅1])
p(λ)=determinant([01-1√2]+[-λ00-λ⋅1])
Step 4.1.2.4
Multiply -1 by 1.
p(λ)=determinant([01-1√2]+[-λ00-λ])
p(λ)=determinant([01-1√2]+[-λ00-λ])
p(λ)=determinant([01-1√2]+[-λ00-λ])
Step 4.2
Add the corresponding elements.
p(λ)=determinant[0-λ1+0-1+0√2-λ]
Step 4.3
Simplify each element.
Step 4.3.1
Subtract λ from 0.
p(λ)=determinant[-λ1+0-1+0√2-λ]
Step 4.3.2
Add 1 and 0.
p(λ)=determinant[-λ1-1+0√2-λ]
Step 4.3.3
Add -1 and 0.
p(λ)=determinant[-λ1-1√2-λ]
p(λ)=determinant[-λ1-1√2-λ]
p(λ)=determinant[-λ1-1√2-λ]
Step 5
Step 5.1
The determinant of a 2×2 matrix can be found using the formula |abcd|=ad-cb.
p(λ)=-λ(√2-λ)-(-1⋅1)
Step 5.2
Simplify the determinant.
Step 5.2.1
Simplify each term.
Step 5.2.1.1
Apply the distributive property.
p(λ)=-λ√2-λ(-λ)-(-1⋅1)
Step 5.2.1.2
Rewrite using the commutative property of multiplication.
p(λ)=-λ√2-1⋅-1λ⋅λ-(-1⋅1)
Step 5.2.1.3
Simplify each term.
Step 5.2.1.3.1
Multiply λ by λ by adding the exponents.
Step 5.2.1.3.1.1
Move λ.
p(λ)=-λ√2-1⋅-1(λ⋅λ)-(-1⋅1)
Step 5.2.1.3.1.2
Multiply λ by λ.
p(λ)=-λ√2-1⋅-1λ2-(-1⋅1)
p(λ)=-λ√2-1⋅-1λ2-(-1⋅1)
Step 5.2.1.3.2
Multiply -1 by -1.
p(λ)=-λ√2+1λ2-(-1⋅1)
Step 5.2.1.3.3
Multiply λ2 by 1.
p(λ)=-λ√2+λ2-(-1⋅1)
p(λ)=-λ√2+λ2-(-1⋅1)
Step 5.2.1.4
Multiply -(-1⋅1).
Step 5.2.1.4.1
Multiply -1 by 1.
p(λ)=-λ√2+λ2--1
Step 5.2.1.4.2
Multiply -1 by -1.
p(λ)=-λ√2+λ2+1
p(λ)=-λ√2+λ2+1
p(λ)=-λ√2+λ2+1
Step 5.2.2
Reorder -λ√2 and λ2.
p(λ)=λ2-λ√2+1
p(λ)=λ2-λ√2+1
p(λ)=λ2-λ√2+1
Step 6
Set the characteristic polynomial equal to 0 to find the eigenvalues λ.
λ2-λ√2+1=0
Step 7
Step 7.1
Use the quadratic formula to find the solutions.
-b±√b2-4(ac)2a
Step 7.2
Substitute the values a=1, b=-√2, and c=1 into the quadratic formula and solve for λ.
√2±√(-√2)2-4⋅(1⋅1)2⋅1
Step 7.3
Simplify.
Step 7.3.1
Simplify the numerator.
Step 7.3.1.1
Apply the product rule to -√2.
λ=√2±√(-1)2√22-4⋅1⋅12⋅1
Step 7.3.1.2
Raise -1 to the power of 2.
λ=√2±√1√22-4⋅1⋅12⋅1
Step 7.3.1.3
Multiply √22 by 1.
λ=√2±√√22-4⋅1⋅12⋅1
Step 7.3.1.4
Rewrite √22 as 2.
Step 7.3.1.4.1
Use n√ax=axn to rewrite √2 as 212.
λ=√2±√(212)2-4⋅1⋅12⋅1
Step 7.3.1.4.2
Apply the power rule and multiply exponents, (am)n=amn.
λ=√2±√212⋅2-4⋅1⋅12⋅1
Step 7.3.1.4.3
Combine 12 and 2.
λ=√2±√222-4⋅1⋅12⋅1
Step 7.3.1.4.4
Cancel the common factor of 2.
Step 7.3.1.4.4.1
Cancel the common factor.
λ=√2±√222-4⋅1⋅12⋅1
Step 7.3.1.4.4.2
Rewrite the expression.
λ=√2±√2-4⋅1⋅12⋅1
λ=√2±√2-4⋅1⋅12⋅1
Step 7.3.1.4.5
Evaluate the exponent.
λ=√2±√2-4⋅1⋅12⋅1
λ=√2±√2-4⋅1⋅12⋅1
Step 7.3.1.5
Multiply -4⋅1⋅1.
Step 7.3.1.5.1
Multiply -4 by 1.
λ=√2±√2-4⋅12⋅1
Step 7.3.1.5.2
Multiply -4 by 1.
λ=√2±√2-42⋅1
λ=√2±√2-42⋅1
Step 7.3.1.6
Subtract 4 from 2.
λ=√2±√-22⋅1
Step 7.3.1.7
Rewrite -2 as -1(2).
λ=√2±√-1⋅22⋅1
Step 7.3.1.8
Rewrite √-1(2) as √-1⋅√2.
λ=√2±√-1⋅√22⋅1
Step 7.3.1.9
Rewrite √-1 as i.
λ=√2±i√22⋅1
λ=√2±i√22⋅1
Step 7.3.2
Multiply 2 by 1.
λ=√2±i√22
λ=√2±i√22
Step 7.4
The final answer is the combination of both solutions.
λ=√2+i√22,√2-i√22
λ=√2+i√22,√2-i√22