Multiply $Ax=b$ by $Q^T$ to get the equivalent triangular systemFor nonsingular $A$, compute $c=Q^{T}b$ and use back substitution:Orthogonal transformations preserve the Euclidean norm, making this a numerically stable reduction. The factors can be applied as stored Householder transformations without explicitly forming $Q$.

Since $Q$ is invertible, $A$ is singular exactly when $R$ is singular. For an upper triangular matrix,Thus a singular $A$ forces at least one zero diagonal entry in $R$. In that case, ordinary division-based back substitution must be replaced by a consistency/rank analysis: the system may have no solution or infinitely many solutions, and a rank-revealing factorization can identify the free parameters.

Apply part (b) to the first column of $A$, using the identity if it is already the zero vector. This gives an orthogonal matrix $H_1$ for which $H_{1}A$ has zeros below its first diagonal entry. At stage $j$, apply a Householder transformation to rows $j,\dots ,n$ of column $j$, and extend it by the identity on the first $j-1$ coordinates. If that trailing column is zero, again use the identity. The transformation preserves zeros in earlier columns, since their trailing entries already vanish.

After at most $n-1$ stages,is an upper triangular matrix. Each $H_j$ is symmetric and orthogonal, so withwe have $Q^{T}Q=I$ andNo invertibility assumption was used, so the Householder QR decomposition also exists for singular matrices.

Let $r=\Vert u{\Vert }_2>0$, and choose $s=1$ if $u_1\ge 0$ and $s=-1$ otherwise. SetThis vector is nonzero, because $w_1$ has magnitude $|u_1|+r$. Direct computation givesso the Householder transformation satisfiesThe first entry is nonzero and every other entry is zero. The sign choice also avoids cancellation when forming the first entry of $w$, which is useful in numerical computation.

For $w\ne 0$, the Householder transformation isLet $P=w{w}^T/(w^{T}w)$. Then $P^T=P$ and $P^2=P$, since $P$ is the orthogonal projection matrix onto the span of $w$. Hence $H^T=H$ andThus $H$ is a symmetric matrix and an orthogonal matrix. Geometrically it reverses the component parallel to $w$ and fixes the perpendicular hyperplane.

By part (c), the best polynomial of degree at most $2$ is odd, hence has the form $p_2(x)=cx$. The orthogonal projection condition with $q=x$ givesThe residual is odd, so its weighted inner products with the even polynomials $1$ and $x^2$ vanish automatically. Thus all three normal equations hold and

Define the reflected polynomial $\tilde{p}(x)=-p(-x)$. It belongs to ${\Pi }_n$. Since $u$ is odd, a change of variable and the evenness of the weight giveThus $\tilde{p}$ is another polynomial least-squares approximation. By uniqueness, $\tilde{p}=p$, soThe conclusion includes the zero polynomial, which is both odd and even.

If $p$ is a minimizer, then for each $q\in {\Pi }_n$ the quadratichas a minimum at $t=0$, so $(u-p,q)=0$. Conversely, if this orthogonality holds, then for every $q\in {\Pi }_n$ the Pythagorean theorem in the inner product space givesThus the minimizers are exactly the orthogonal projections characterized byFor existence, write $p={\sum }_{j=0}^n{c}_j{x}^j$. The conditions form the normal equations $Gc=b$, with Gram matrix $G_{ij}=(x^i,x^j)$ and $b_i=(u,x^i)$. For every nonzero coefficient vector $v$,Indeed a nonzero polynomial is nonzero on some open interval away from $0$, where the weight is positive. Hence $G$ is a positive-definite matrix and is invertible, proving existence and uniqueness of $c$. Uniqueness also follows from the preceding Pythagorean theorem identity. The zero of the weight at one point therefore causes no degeneracy on continuous functions or polynomials.

A polynomial least-squares approximation $p_n\in {\Pi }_n$ minimizes the weighted residual norm:Equivalently it minimizes ${\int }_{-1}^1{x}^2(u(x)-q(x){)}^{2}dx$. This is a best approximation within a finite-dimensional polynomial space, without an interpolation requirement.

Let $v_i=[(X^{T}X{)}^{-1}{]}_{ii}>0$. Part (b) givesThe residual vector and coefficient estimator are independent: they are jointly normal and their cross-covariance matrix is zero, since $X^T(I-H)=0$. Hence $Z_i$ is independent of $U=(n-p)s^2/{\sigma }^2\sim {\chi }_{n-p}^2$. By the definition of Student's t-distribution,Let $t_{n-p,1-\alpha /2}$ be the $1-\alpha /2$ quantile of this distribution. Inverting the central probability statement gives the exact marginal confidence intervalIts coverage is $1-\alpha$ under the stated Gaussian linear model; simultaneous coverage for all coefficients would require a separate adjustment.

Use the residual mean squareTo justify its denominator, let $H=X(X^{T}X{)}^{-1}{X}^T$, the orthogonal projection matrix onto the column space of $X$. Then the residual is $(I-H)ϵ$. The idempotent matrix $I-H$ has rank $n-p$, so diagonalizing it with an orthogonal matrix givesConsequently $\mathbb{E}{s}^2={\sigma }^2$, making $s^2$ an unbiased estimator. If specifically asked for maximum likelihood estimation, maximizing over ${\sigma }^2$ gives the alternative ${\hat{\sigma }}_{\mathrm{ML}}^2=\Vert Y-X\hat{\theta }{\Vert }^2/n$, which is biased downward by the factor $(n-p)/n$.

Write $A=(X^{T}X{)}^{-1}{X}^T$. Since $AX=I_p$,The errors have mean zero and covariance matrix ${\sigma }^2{I}_n$. ThereforeThus $\hat{\theta }$ is an unbiased estimator. In addition, the multivariate normal distribution is preserved by linear maps, so $\hat{\theta }$ is multivariate normal with the displayed mean and covariance matrix.

The Gaussian linear model has log likelihood function, up to an additive constant,For each fixed positive ${\sigma }^2$, maximize this by minimizing the residual sum of squares. Differentiation gives the normal equationsThe full column rank of $X$ makes $X^{T}X$ positive definite, hence invertible, soThe quadratic objective is strictly convex, which proves uniqueness. The joint maximum-likelihood estimate has the same coefficient estimate, except on the probability-zero event of an exactly zero residual when the positive-variance likelihood has no finite joint maximizer.

The log likelihood ratio statistic comparing the unrestricted fit with the independence fit isThe unrestricted model has $rc-1=8$ free parameters. Independence has $(r-1)+(c-1)=4$, so the difference is $(r-1)(c-1)=4$. By Wilks theorem, under the null hypothesis the statistic has asymptotically a chi-squared distribution with $4$ degrees of freedom, rather than $6$, $8$, or $9$.

For the supplied statistic, the approximate p-value is $1-F_4(11.8)=0.02$. Thus reject independence at the 5% level, but not at the 1% level: the data provide evidence of an association. The approximation is reasonable here; the smallest fitted cell count is $27\cdot 99/299\approx 8.94$, so all fitted counts exceed $5$. The statistical hypothesis test concerns association, not a causal effect of personality.

Under independence, the log likelihood function separates:Maximizing the two terms over their respective probability simplices, by the same Lagrange multiplier argument as in part (a), givesTherefore the constrained maximum-likelihood estimates areThe associated fitted cell counts are $E_{ij}=N{\hat{\theta }}_{ij}^0=n_{i.}{n}_{.j}/N$.

Ignoring a factor independent of the parameters, the likelihood function isSince every count is positive, a boundary value ${\theta }_{ij}=0$ has zero likelihood. At an interior optimum, a Lagrange multiplier for $\sum {\theta }_{ij}=1$ gives $n_{ij}/{\theta }_{ij}=\lambda$. Summing the resulting ${\theta }_{ij}=n_{ij}/\lambda$ gives $\lambda =N$, soThe log likelihood function is strictly concave on the positive simplex, because its Hessian matrix is diagonal with entries $-n_{ij}/{\theta }_{ij}^2<0$. Hence this critical point is the unique global maximum-likelihood estimate.